1.
o
Arithmetic Progression And
Geometric Progression
Introduction : :
We have seen different types of numbers since our childhood. Like set of
positive even numbers, odd numbers etc. In all these sets some specific
pattern is followed. In nature also we see that some pattern is followed e.g.
Arrangement of petals in flower or arrangement in fruits or in tree.
All these arrangements in nature look beautiful because some pattern is
followed in all of them.
o
Sequence :
A sequence is a collection of numbers arranged in a definite order according
to some definite rule.
Each number in the sequence is called a term of the sequence.
The number in the first position is called the first term and it is denoted by t1.
Similarly the number in the second and third positions are denoted as t2
and t3 respectively.
In general, the number in the nth position is called nth term and is denoted
by tn. A sequence is usually denoted by { tn } or < tn > and read as sequence tn.
o
Examples of sequences :
1.
2.
3.
4.
2, 4, 6, 8, ......
3, 6, 9, 12, ......
5, 25, 125, 625, ......
– 4, – 2, 0, 2, ......
1 1 1
1
, ,
,
, ......
2 6 18 54
5.
o
Here
Here
Here
Here
t1
t1
t1
t1
=
=
=
=
2, t2 = 4, t3 = 6, t4 = 8, .......
3, t2 = 6, t3 = 9, t4 = 12, .......
5, t2 = 25, t3 = 125, t4 = 625, .......
– 4, t2 = – 2, t3 = 0, t4 = 2, .......
1
1
1
1
Here t1 =
,t =
, t =
,t =
, .......
2 2
6 3 18 4
54
Sum of first n terms of a sequence :
S1 = t1
S2 = t1 + t2
S3 = t1 + t2 + t3
S4 = t1 + t2 + t3 + t4
In general Sn = t1 + t2 + t3 + ....... + tn.
From the above equations we observed that
S1 = t1
S2 – S1 = t2
S3 – S2 = t3
In general Sn – Sn–1 = tn.
 tn = Sn – Sn–1.
S C H O O L S E C TI O N
1
MT
ALGEBRA
o
EDUCARE LTD.
Types of sequence :
If the number of terms in a sequence is finite then it is called a finite
sequence. e.g. : 5, 9, 13, 17.
If the number of terms in the sequence is not finite then it is called
infinite sequence. e.g. : 7, 10, 13, 16, ......
EXERCISE - 1.1 (TEXT BOOK PAGE NO. 4) :
1.
(i)
Sol.
For each sequence,
1, 2, 4, 7, 11, .....
t1 = 1 + 0
=
t2 = 1 + 1
=
t3 = 2 + 2
=
=
t4 = 4 + 3
t5 = 7 + 4
=
t 6 = 11 + 5 =
t 7 = 16 + 6 =
t 8 = 22 + 7 =
t 9 = 29 + 8 =
find the next four terms.
(1 mark)
1
2
4
7
11
16
22
29
37
 The next four terms of the sequence are 16, 22, 29 and 37.
(ii)
Sol.
3, 9, 27,
t1 =
t2 =
t3 =
t4 =
t5 =
t6 =
t7 =
t8 =
(1 mark)
81, .....
31
= 3
2
3
= 9
33
= 27
34
= 81
35
= 243
36
= 729
7
3
= 2187
38
= 6561
 The next four terms of the sequence are 243, 729, 2187 and 6561.
(iii)
Sol.
1, 3, 7, 15, 31, ....
t 1 = 0 + 20
=
1
t2 = 1 + 2
=
t 3 = 3 + 22
=
t 4 = 7 + 23
=
t 5 = 15 + 24 =
t 6 = 31 + 25 =
t 7 = 63 + 26 =
t 8 = 127 + 27 =
t 9 = 255 + 28 =
(1 mark)
0+1
1+2
3+4
7+8
15 + 16
31 + 32
63 + 64
127 + 128
255 + 256
=
=
=
=
=
=
=
=
=
1
3
7
15
31
63
127
255
511
 The next four terms of the sequence are 63, 127, 255 and 511.
(iv)
Sol.
2
192, – 96, 48, – 24, ....
t 1 = 192
192
t2 = – 2
= – 96
t3
=
t4
=
– 96
–2
48
–2
(1 mark)
= 48
= – 24
S C H O O L S E C TI O N
MT
ALGEBRA
EDUCARE LTD.
t5
=
– 24
–2
= 12
t6
=
12
–2
= –6
t7
=
–6
–2
= 3
t8
=
3
–2
=
–
3
2
 The next four terms of the sequence are 12, – 6, 3 and –
(v)
Sol.
2, 6, 12,
t1 =
t2 =
t3 =
t4 =
t5 =
t6 =
t7 =
t8 =
t9 =
20, 30, ...
0+2
=
2+4
=
6+6
=
12 + 8 =
20 + 10 =
30 + 12 =
42 + 14 =
56 + 16 =
72 + 18 =
3
.
2
(1 mark)
2
6
12
20
30
42
56
72
90
 The next four terms of the sequence are 42, 56, 72 and 90.
(vi)
Sol.
0.1, 0.01, 0.001, 0.0001, ....
t1 =
0.1
t2
=
0.1
10
= 0.01
t3
=
0.01
10
= 0.001
t4
=
0.001
10
= 0.0001
t5
=
0.0001
10
= 0.00001
t6
=
0.00001
10
= 0.000001
t7
=
0.000001
10
= 0.0000001
t8
=
0.0000001
= 0.00000001
10
(1 mark)
 The next four terms of the sequence are 0.00001, 0.000001,
0.0000001 and 0.00000001.
(vii)
Sol.
2, 5, 8, 11,
t1 = 2
t2 = 2
t3 = 5
t4 = 8
S C H O O L S E C TI O N
(1 mark)
....
+3
+3
+3
= 5
= 8
= 11
3
MT
ALGEBRA
t5
t6
t7
t8
=
=
=
=
11
14
17
20
+
+
+
+
3
3
3
3
=
=
=
=
EDUCARE LTD.
14
17
20
23
 The next four terms of the sequence are 14, 17, 20 and 23.
(viii)
Sol.
– 25, – 23, – 21, – 19, .....
t 1 = – 25
t 2 = – 25 + 2 = – 23
t 3 = – 23 + 2 = – 21
t 4 = – 21 + 2 = – 19
t 5 = – 19 + 2 = – 17
t 6 = – 17 + 2 = – 15
t 7 = – 15 + 2 = – 13
t 8 = – 13 + 2 = – 11
(1 mark)
 The next four terms of the sequence are – 17, – 15, – 13 and – 11.
(ix)
Sol.
2, 4, 8, 16, .....
t 1 = 21
t 2 = 22
t 3 = 23
t 4 = 24
t 5 = 25
t 6 = 26
t 7 = 27
t 8 = 28
(1 mark)
=
=
=
=
=
=
=
=
2
4
8
16
32
64
128
256
 The next four terms of the sequence are 32, 64, 128 and 256.
(x)
Sol.
1 1 1
1
,
,
,
, ....
2 6 18 54
1
2
1 1

2 3
1 1

6 3
(2 mark)
t1
=
t2
=
t3
=
t4
=
t5
=
t6
=
t7
=
1
1

486 3
=
1
1458
t8
=
1
1

1458 3
=
1
4374
1
1

18 3
1
1

54 3
1
1

162 3
=
=
=
=
=
1
6
1
18
1
54
1
162
1
486
 The next four terms of the sequence are
4
1
1
1
1
,
,
and
.
162 486 1458
4374
S C H O O L S E C TI O N
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ALGEBRA
EDUCARE LTD.
EXERCISE - 1.1 (TEXT BOOK PAGE NO. 4) :
2.
(i)
Sol.
Find the first
are given :
tn = 4n – 3
t n = 4n –
 t 1 = 4 (1)
 t 2 = 4 (2)
 t 3 = 4 (3)
 t 4 = 4 (4)
 t 5 = 4 (5)
five terms of the following sequences, whose ‘nth’ terms
(1 mark)
3
–
–
–
–
–
3
3
3
3
3
=
=
=
=
=
4–3
8–3
12 – 3
16 – 3
20 – 3
=
=
=
=
=
1
5
9
13
17
 The first five terms of the sequence are 1, 5, 9, 13 and 17.
(ii)
Sol.
tn = 2n –
tn =
 t1 =
 t2 =
 t3 =
 t4 =
 t5 =
5
2n –
2 (1)
2 (2)
2 (3)
2 (4)
2 (5)
(1 mark)
5
–
–
–
–
–
5
5
5
5
5
=
=
=
=
=
2–5
4–5
6–5
8–5
10 – 5
=
=
=
=
=
–3
–1
1
3
5
 The first five terms of the sequence are – 3, – 1, 1, 3 and 5.
(iii)
Sol.
tn = n
tn
 t1
 t2
 t3
 t4
 t5
+2
=
=
=
=
=
=
(1 mark)
n+2
1+2
2+2
3+2
4+2
5+2
=
=
=
=
=
3
4
5
6
7
 The first five terms of the sequence are 3, 4, 5, 6 and 7.
(iv)
Sol.
tn = n 2
tn
 t1
 t2
 t3
 t4
 t5
–
=
=
=
=
=
=
2n
n2 – 2n
1 – 2 (1)
22 – 2 (2)
32 – 2 (3)
42 – 2 (4)
52 – 2 (5)
(1 mark)
=
=
=
=
=
1–2 =
4–4 =
9–6 =
16 – 8 =
25 – 10 =
–1
0
3
8
15
 The first five terms of the sequence are – 1, 0, 3, 8 and 15.
(v)
Sol.
tn = n 3
tn
 t1
 t2
 t3
 t4
 t5
(1 mark)
=
=
=
=
=
=
n3
13
23
33
43
53
=
=
=
=
=
1
8
27
64
125
 The first five terms of the sequence are 1, 8, 27, 64 and 125.
(vi)
Sol.
tn =
1
n 1
tn
=
S C H O O L S E C TI O N
(1 mark)
1
n 1
5
MT
ALGEBRA
 t1
=
 t2
=
 t3
=
 t4
=
 t5
=
1
11
1
2 1
1
3 1
1
4 1
1
5 1
=
=
=
=
=
EDUCARE LTD.
1
2
1
3
1
4
1
5
1
6
 The first five terms of the sequence are
1 1 1 1
1
,
,
,
and
.
2 3 4 5
6
EXERCISE - 1.1 (TEXT BOOK PAGE NO. 4) :
3.
(i)
Sol.
Find the first three terms of the sequences for which Sn is given below :
Sn = n2 (n + 1)
(2 marks)
S n = n2 (n + 1)
 S 1 = 12 (1 + 1) = 1 (2)
= 2
 S 2 = 22 (2 + 1) = 4 (3)
= 12
2
 S 3 = 3 (3 + 1) = 9 (4)
= 36
We know that,
t1 = S1
= 2
t 2 = S2 – S1
= 12 – 2 = 10
t 3 = S3 – S2
= 36 – 12 = 24
 The first three terms of the sequence are 2, 10 and 24.
(ii)
Sol.
2
2
Sn = n (n  1)
4
n2 (n  1)2
Sn =
4
(2 marks)
1 4
4
 S1 =
12 (1  1)2
4
 S2 =
22 (2  1)2
4 (3)2
=
= 9
4
4
32 (3  1)2
4
We know that,
t1 = S1
t 2 = S2 – S1
t 3 = S3 – S2
 S3 =
=
=
1 (2)2
4
=
9  42
= 9×4
4
= 1
= 36
= 1
= 9–1 = 8
= 36 – 9 = 27
 The first three terms of the sequence are 1, 8 and 27.
(iii)
Sol.
6
n (n  1) (2n  1)
6
Sn =
n (n  1) (2n  1)
6
 S1 =
1 (1  1) [2 (1)  1]
6
(2 marks)
=
1 2  3
6
=
6
6
= 1
S C H O O L S E C TI O N
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ALGEBRA
EDUCARE LTD.
2 (2  1) [2 (2)  1]
=
6
3 (3  1) [2 (3)  1]
 S3 =
=
6
We know that,
t1 = S1
= 1
t 2 = S2 – S1 = 5 – 1 =
t 3 = S3 – S2 = 14 – 5 =
 S2 =
235
6
347
6
=
30
= 5
6
= 14
4
9
 The first three terms of the sequence are 1, 4 and 9.
o
Progression :
Whenever we write some numbers one after the other, keeping a fix relation
between two consecutive terms then we say the numbers are in progression.
e.g. 1, 5, 10, 19, ..........
o
Arithmetic Progression :
Arithmetic Progression is a sequence in which the difference between two
consecutive terms is a constant.
e.g. 4, 8, 12, 16, ...........
Here, the difference between any two consecutive terms is 4 which is a
constant. Therefore, the sequence is an A.P.
e.g. 2, 4, 8, 16, ...........
Here, the difference between any two consecutive terms is not a constant.
Therefore, the sequence is not an A.P.
EXERCISE - 1.2 (TEXT BOOK PAGE NO. 8) :
1.
(i)
Sol.
Which of the following lists of numbers are Arithmetic Progressions?
Justify.
1, 3, 6, 10, .....
(1 mark)
t1 = 1, t2 = 3, t3 = 6, t4 = 10
t2 – t 1 = 3 – 1
= 2
t3 – t 2 = 6 – 3
= 3
t4 – t3 = 10 – 6 = 4
 The difference between two consecutive terms is not constant.
 The sequence is not an A.P.
(ii)
Sol.
3, 5, 7, 9, 11, .....
(1 mark)
t1 = 3, t2 = 5, t3 = 7, t4 = 9, t5 = 11
t2 – t 1 = 5 – 3
= 2
t3 – t 2 = 7 – 5
= 2
t4 – t 3 = 9 – 7
= 2
t5 – t4 = 11 – 9 = 2
 The difference between two consecutive terms is 2 which is constant.
 The sequence is an A.P.
(iii)
Sol.
1, 4, 7, 10, ....
(1 mark)
t1 = 1, t2 = 4, t3 = 7, t4 = 10
t2 – t 1 = 4 – 1
= 3
t3 – t 2 = 7 – 4
= 3
t4 – t3 = 10 – 7 = 3
 The difference between two consecutive terms 3 which is constant.
 The sequence is an A.P.
S C H O O L S E C TI O N
7
ALGEBRA
MT
EDUCARE LTD.
(iv)
Sol.
3, 6, 12, 24, ....
(1 mark)
t1 = 3, t2 = 6, t3 = 12, t4 = 24
= 3
t2 – t 1 = 6 – 3
t3 – t2 = 12 – 6 = 6
t4 – t3 = 24 – 12 = 12
 The difference between two consecutive terms is not constant.
 The sequence is not an A.P.
(v)
Sol.
22, 26, 28, 31, ...
(1 mark)
t1 = 22, t2 = 26, t3 = 28, t4 = 31
t2 – t1 = 26 – 22 = 4
t3 – t2 = 28 – 26 = 2
t4 – t3 = 21 – 28 = 3
 The difference between two consecutive terms is not constant.
 The sequence is not an A.P.
(vi)
Sol.
(1 mark)
0.5, 2, 3.5, 5, ...
t1 = 0.5, t2 = 2, t3 = 3.5, t4 = 5
t2 – t1 = 2 – 0.5 = 1.5
t3 – t2 = 3.5 – 2 = 1.5
t4 – t3 = 5 – 3.5 = 1.5
 The difference between two consecutive terms is 1.5 which is constant.
 The sequence is an A.P.
(vii)
Sol.
4, 3, 2, 1, ....
(1 mark)
t1 = 4, t2 = 3, t3 = 2, t4 = 1,
= –1
t2 – t 1 = 3 – 4
= –1
t3 – t 2 = 2 – 3
= –1
t4 – t 3 = 1 – 2
 The difference between two consecutive terms is –1 which is constant.
 The sequence is an A.P.
(viii)
Sol.
– 10, – 13, – 16, – 19, .....
t1 = – 10, t2 = – 13, t3 = – 16, t4 = – 19
t2 – t1 = – 13 – (– 10) = – 13 + 10 =
t3 – t2 = – 16 – (– 13) = – 16 + 13 =
t4 – t3 = – 19 – (– 16) = – 19 + 16 =
 The difference between two consecutive
 The sequence is an A.P.
(1 mark)
–3
–3
–3
terms is – 3 which is constant.
EXERCISE - 1.2 (TEXT BOOK PAGE NO. 8) :
2.
(i)
Sol.
Write the first five terms of the following Arithmetic Progressions where,
the common difference ‘d’ and the first term ‘a’ are given :
a = 2, d = 2.5
(1 mark)
a = 2, d = 2.5
Here, t 1 = a
=
2
t 2 = t1 + d =
2 + 2.5
= 4.5
t 3 = t2 + d =
4.5 + 2.5 = 7
t 4 = t3 + d =
7 + 2.5
= 9.5
t 5 = t4 + d =
9.5 + 2.5 = 12
 The first five terms of the A.P. are 2, 4.5, 7, 9.5 and 12.
8
S C H O O L S E C TI O N
MT
(ii)
Sol.
ALGEBRA
EDUCARE LTD.
a = 10, d = – 3
a = 10, d = – 3
Here, t 1 = a
t 2 = t1
t 3 = t2
t 4 = t3
t 5 = t4
(1 mark)
+
+
+
+
d
d
d
d
=
=
=
=
=
10
10 + (– 3)
7 + (– 3)
4 + (– 3)
1 + (– 3)
=
=
=
=
10 – 3
7–3
4–3
1–3
=
=
=
=
7
4
1
–2
 The first five terms of the A.P. are 10, 7, 4, 1 and – 2.
(iii)
Sol.
a = 4, d = 0
a = 4, d = 0
Here, t 1 =
t2 =
t3 =
t4 =
t5 =
(1 mark)
a
t1
t2
t3
t4
+
+
+
+
d
d
d
d
=
=
=
=
=
4
4
4
4
4
+
+
+
+
0
0
0
0
=
=
=
=
4
4
4
4
 The first five terms of the A.P. are 4, 4, 4, 4 and 4.
(iv)
Sol.
a = 5, d = 2
a = 5, d = 2
Here, t 1 =
t2 =
t3 =
t4 =
t5 =
(1 mark)
a
t1
t2
t3
t4
+
+
+
+
d
d
d
d
=
=
=
=
=
5
5+2
7+2
9+2
11 + 2
=
=
=
=
7
9
11
13
 The first five terms of the A.P. are 5, 7, 9, 11 and 13.
(v)
Sol.
a = 3, d = 4
a = 3, d = 4
Here, t 1 =
t2 =
t3 =
t4 =
t5 =
(1 mark)
a
t1
t2
t3
t4
+
+
+
+
d
d
d
d
=
=
=
=
=
3
3+4
7+4
11 + 4
15 + 4
=
=
=
=
7
11
15
19
 The first five terms of the A.P. are 3, 7, 11, 15 and 19.
(vi)
Sol.
a = 6, d = 6
a = 6, d = 6
Here, t 1 =
t2 =
t3 =
t4 =
t5 =
(1 mark)
a
t1
t2
t3
t4
+
+
+
+
d
d
d
d
=
=
=
=
=
6
6+6
12 + 6
18 + 6
24 + 6
=
=
=
=
12
18
24
30
 The first five terms of A.P. are 6, 12, 18, 24 and 30.
NOTE : In an A.P. the difference between two consecutive terms is
constant and it is denoted as ‘d’ and first term of an A.P. is
denoted as ‘a’.
S C H O O L S E C TI O N
9
MT
ALGEBRA
o
General term of an A.P. (tn) :
t1 = a
= a + 0d
= a + 1d
t2 = a + d
t3 = a + d + d
= a + 2d
= a + 3d
t4 = a + d + d + d
t5 = a + d + d + d + d = a + 4d
In general nth term, of an A.P.





t1 = a +
t2 = a +
t3 = a +
t4 = a +
t5 = a +
tn = a +
EDUCARE LTD.
(1 – 1) d
(2 – 1) d
(3 – 1) d
(4 – 1) d
(5 – 1) d
(n – 1) d
EXERCISE - 1.3 (TEXT BOOK PAGE NO. 12) :
1.
Sol.
Find the twenty fifth term of the A. P. : 12, 16, 20, 24, .....
For the given A.P. 12, 16, 20, 24, .....
Here, a = t1 = 12
d = t2 – t1 = 16 – 12 = 4
We know,
t n = a + (n – 1) d
 t 25 = a + (25 – 1) d
 t 25 = 12 + 24 (4)
 t 25 = 12 + 96
 t 25 = 108
(2 marks)
 The twenty fifth term of A.P. is 108.
EXERCISE - 1.3 (TEXT BOOK PAGE NO. 12) :
2.
Sol.
(2 marks)
Find the eighteenth term of the A. P. : 1, 7, 13, 19, .....
For the given A.P. 1, 7, 13, 19, .....
Here, a = t1 = 1
d = t 2 – t1 = 7 – 1 = 6
We know,
t n = a + (n – 1) d
 t 18 = a + (18 – 1) d
 t 18 = 1 + 17 (6)
 t 18 = 1 + 102
 t 18 = 103
 Eighteenth term of A.P. is 103.
PROBLEM SET - 1 (TEXT BOOK PAGE NO. 162) :
1.
Sol.
(2 marks)
Find t11 from the following A.P. 4, 9, 14, ..... .
For the A.P. 4, 9, 14, .....
a = 4, d = 5
tn
= a + (n – 1) d
 t 11
= 4 + (11 – 1) 5
 t 11
= 4 + 50
 t 11
= 54
EXERCISE - 1.3 (TEXT BOOK PAGE NO. 12) :
3.
Find tn for an Arithmetic Progression where t3 =
Given : For an A.P. t3 = 22 and t17 = – 20
Find :
t n.
Sol.
t n = a + (n – 1) d
t 3 = a + (3 – 1) d
22 = a + 2d
 a + 2d = 22
......(i)
t 17 = a + (17 – 1) d
– 20 = a + 16d
 a + 16d = – 20
......(ii)
10
22, t17 = – 20.
(3 marks)
S C H O O L S E C TI O N
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



ALGEBRA
Subtracting (ii) from (i),
a + 2d =
22
a + 16d = – 20
(–) (–)
(+)
– 14d =
42
42
d = –14
d =
–3
Substituting d = – 3 in (i),
a + 2 (– 3) = 22
a – 6 = 22
a = 22 + 6
a = 28
tn
 tn
 tn
= a + (n – 1) d
= 28 + (n – 1) (– 3)
= 28 – 3n + 3
 tn
= 31 – 3n
EXERCISE - 1.3 (TEXT BOOK PAGE NO. 12) :
4.
For an A. P. if t4 = 12, and d = – 10, then find its general term.
Given : For an A.P. t4 = 12, d = – 10
Find :
General term { tn }
Sol.
t n = a + (n – 1) d
 t 4 = a + (4 – 1) d
 12 = a + 3 (– 10)
 a
= 12 + 30
 a
= 42
t n = a + (n – 1) d
 t n = 42 + (n – 1) (– 10)
 t n = 42 – 10n + 10
 t n = 52 – 10n
(3 marks)
 The general term of A.P. is 52 – 10n.
EXERCISE - 1.3 (TEXT BOOK PAGE NO. 12) :
5.
Sol.
Given the following sequence, determine whether it is arithmetic or
not. If it is an Arithmetic Progression, find its general term :
(3 marks)
– 5, 2, 9, 16, 23, 30, .....
– 5, 2, 9, 16, 23, 30, .....
t1 = – 5, t2 = 2, t3 = 9, t4 = 16, t5 = 23, t6 = 30
t2 – t1 = 2 – (– 5) = 2 + 5 = 7
t3 – t 2 = 9 – 2
= 7
t4 – t3 = 16 – 9 = 7
t5 – t4 = 23 – 16 = 7
t6 – t5 = 30 – 23 = 7
 The difference between two consecutive terms is 7 which is a constant.
 The sequence is an A.P. with a = t1 = – 5.
Common difference (d) = 7
tn = a + (n – 1) d
tn = – 5 + (n – 1) 7
tn = – 5 + 7n – 7
tn = 7n – 12
 The general term of A.P. is 7n – 12.
S C H O O L S E C TI O N
11
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EXERCISE - 1.3 (TEXT BOOK PAGE NO. 12) :
6.
Sol.
Given the following sequence, determine if it is arithmetic or not. If it
is an Arithmetic Progression, find its general term.
(2 marks)
5, 2, – 2, – 6, – 11, .....
5, 2, – 2, – 6, – 11, .....
t1 = 5, t2 = 2, t3 = – 2, t4 = – 6, t5 = – 11
t2 – t 1 = 2 – 5
= –3
t3 – t 2 = – 2 – 2 = – 4
t4 – t3 = – 6 – (– 2) = – 6 + 2 = – 4
t5 – t4 = – 11 – (– 6) = – 11 + 6 = – 5
 The difference between two consecutive terms is not a constant.
 The sequence is not an A.P.
EXERCISE - 1.3 (TEXT BOOK PAGE NO. 12) :
7.
Sol.
How many three digit natural numbers are divisible by 4 ?
(3 marks)
The three digit natural numbers that are divisible by 4 are as follows
100, 104, 108, ........ 996.
These numbers form an A.P. with a = t1 = 100,
d = t2 – t1 = 104 – 100 = 4.
Let, tn = 996
We know that for an A.P.
t n = a + (n – 1) d
 996 = 100 + (n – 1) 4
 996 = 100 + 4n – 4
 996 = 96 + 4n
 4n = 996 – 96
 4n = 900
900
 n =
4
 n = 225
 There are 225 three digit natural numbers that are divisible by 4.
PROBLEM SET - 1 (TEXT BOOK PAGE NO. 162) :
7.
Sol.
The sum of first n terms of an A.P. is 3n + n2 ten (i) find first term and
sum of first two terms. (ii) find second, third and 15th term. (3 marks)
Sn
= 3n + n2
S1
= 3(1) + (1)2
 S1
= 3+1
 S1
= 4
 S1
S2
 S2
= t1 = 4
= 3(2) + (2)2
= 6+4
S2
= 10
t2
 t2
= S2 – S1
= 10 – 4
 t2
= 6
Now,
t1 + t2 = 4 + 6
 t1 + t2 = 10
a
= 4
d
= 6–4=2
12
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t3
 t3
= t2 + d
= 6+2
 t3
= 8




tn
t 15
t 15
t 15
t 15
 t 15
=
=
=
=
=
a
a
a
4
4
+ (n – 1)d
+ (15 – 1)d
+ 14d
+ 14(2)
+ 28
= 32
EXERCISE - 1.3 (TEXT BOOK PAGE NO. 12) :
8.
(i)
Sol.
The 11th term and the 21st term of an A.P. are 16 and 29 respectively, find :
the 1st term and the common difference.
(2 marks)
t n = a + (n – 1) d
t 11 = a + (11 – 1) d
16 = a + 10d
 a + 10d = 16
.......(i)
t 21 = a + (21 – 1) d
29 = a + 20d
 a + 20d = 29
......(ii)
Subtracting (ii) from (i),
a + 10d = 16
a + 20d = 29
(–) (–)
(–)
– 10d = – 13

10d = 13
13

d =
10

d = 1.3
Substituting d = 1.3 in (i),
a + 10 (1.3) = 16

a + 13 = 16

a = 16 – 13

a = 3
 The first term is 3 and the common difference is 1.3
(ii)
Sol.
the 34th term
t n = a + (n – 1) d
t 34 = a + (34 – 1) d
= 3 + 33 (1.3)
= 3 + 42.9
t 34 = 45.9
(1 mark)
 Thirty fourth term of A.P. is 45.9.
(iii)
Sol.
‘n’ such that tn = 55.
tn
= a + (n – 1) d
 55 = 3 + (n – 1) d
 55 = 3 + (n – 1) 1.3
 55 = 3 + 1.3n – 1.3
 55 = 1.7 + 1.3n
 53.3 = 1.3n
S C H O O L S E C TI O N
(1 mark)
13
ALGEBRA
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53.3
= n
1.3
533
 n
=
13
 n
= 41

PROBLEM SET - 1 (TEXT BOOK PAGE NO. 162) :
2.
Sol.
Find first negative term from following A.P. 122, 116, 110, ...... .
(4 marks)
(Note : find smallest n such that tn < 0).
122, 116, 110, .............
But ‘n’ is term number
a = t1 = 122
which is a natural number
d
= t2 – t1 = 116 – 122 = –6
The first natural number
= a + (n – 1) d
tn
greater than 21.33... is
tn
= 122 + (n – 1) (–6)
‘22’
tn
= 122 – 6n + 6
When n = 21
tn
= 128 – 6n
tn
= 128 – 6n
when want smallest n such that
 t 21
= 128 – 6 × 21
tn
< 0
 t 21
= 128 – 126
128 – 6n < 0
128 < 6n
t 21
= 2
Dividing both sides by 6
When n = 22
tn
= 128 – 6n
128

< n
 t 22
= 128 – 6 × 22
6
 t 22
= 128 – 132
 21.33... < n
t 22
= –4
 First negative term of A.P. is – 4.
o
Sum of the first n terms of an A.P. :




To find sum of first n terms of an A.P. we use
Gauss technique.
One day Gauss was asked to find sum of all natural
numbers from 1 to 100. He wrote,
S = 1 + 2 + 3 + ..... + 99 + 100
Again he wrote,
S = 100 + 99 + 98 + ..... + 2 + 1
He added both this statements
2S = (100 + 1) + (99 + 2) + (98 + 3) + .....
+ (2 + 99) + (1 + 100)
2S = 101 + 101 + 101 + ..... + 101 (100 times)
2S = 101 × 100
2S = 10100
S = 5050

14
The same technique we use for finding sum of first n terms of an A.P. in
the following manner.
The sum of the first ‘n’ terms of an A.P. denoted by Sn.
If ‘a’ is the first term and ‘d’ is the common difference, then the terms of
A.P. are a, a + d, a + 2d, ....... a + (n – 1) d, .....
Here, tn = a + (n – 1) d, n  N.
Let us denote tn by l
The first ‘n’ terms of the A.P. are a, a + d, a + 2d, ...... l – 2d, l – d, l
Therefore, sum of the first n terms of the A.P. is
S C H O O L S E C TI O N
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






ALGEBRA
Sn = a + (a + d) + (a + 2d) + .... + (l – 2d) + (l – d) + l
Also,
Sn = l + (l – d) + (l – 2d) + .... + (a + 2d) + (a + d) + a
Adding the corresponding terms of the above equations,
2S n = (a + l) + (a + l) + (a + l) + ...... + (a + l) + (a + l) + (a + l) { n times }
2Sn = n (a + l)
n
Sn =
(a + l)
2
n
Sn =
(t + tn)
.......(i)
[ a = t1 and l = tn]
2 1
Here, t1 = a and tn = [a + (n  1) d]
n
Sn =
[a + { a + (n  1) d }]
2
n
Sn =
[a + a + (n  1) d]
2
n
Sn =
[2a + (n - 1) d]
.......(ii)
2
EXERCISE - 1.4 (TEXT BOOK PAGE NO. 14) :
1.
Sol.
Find the sum of the first n natural numbers and hence find the sum of
(4 marks)
first 20 natural numbers.
The first n natural numbers are 1, 2, 3, 4, ............
These natural number form an A.P. with a = 1, d = 1
n
 Sn =
[2a + (n – 1) d]
Alternative method :
2
n
n
 Sn =
[2 (1) + (n – 1) 1]
Sn =
[t + tn]
2
2 1
n
 Sn =
[2 + n – 1]
20
2
[t1 + t20]
S 20 =
2
n
 Sn =
(n + 1)
= 10 [1 + 20]
2
= 10 [21]
20
 S 20 =
(20 + 1)
 S 20 = 210
2
 S 20 = 10 (21)
 S 20 = 210
 Sum of first twenty natural numbers is 210.
EXERCISE - 1.4 (TEXT BOOK PAGE NO. 14) :
2.
Sol.
Find the sum of all odd natural numbers from 1 to 150.
The odd natural numbers from 1 to 150 are as follows
1, 3, 5, 7, 9, .........., 149.
These numbers form an A.P. with a = 1, d = 2
Let, 149 be nth term of an A.P.
tn
= 149
tn
= a + (n – 1) d
149
= 1 + (n – 1) 2
149
= 1 + 2n – 2
149
= 2n – 1
149 + 1 = 2n
 2n
= 150
 n
= 75
 149 is 75th term of A.P.
S C H O O L S E C TI O N
(4 marks)
15
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ALGEBRA
 We have to find sum of 75 terms i.e. S75
n
Sn
=
[2a + (n – 1) d]
2
75
 S 75
=
[2 (1) + (75 – 1) 2]
2
75
 S 75
=
[2 + 74 [2)]
2
75
[2 + 148]
=
2
75
(150)
=
2
= 75 (75)
 S 75
= 5625
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Alternative method :
Sn =
S 75 =
=
=
=
 S 75 =
n
[t + tn]
2 1
75
[t1 + t75]
2
75
[1 + 149]
2
75
[150]
2
75 [75]
5625
 Sum of all odd natural from 1 to 150 is 5625.
PROBLEM SET - 1 (TEXT BOOK PAGE NO. 162) :
3.
Sol.
Find the sum of first 11 positive numbers which are multiples of 6.
(2 marks)
The positive integers which are divisible by 6 are 6, 12, 18, 24, ........
The number form an A.P. with a = 6, d = 6.
The sum of first 11 positive integers divisible by 6 is (S11)
n
Sn
=
[2a + (n – 1) d]
2
11
 S 11 =
[2a + (11 – 1) d]
2
11
[2 (6) + 10 (6)]
=
2
11
[12 + 60]
=
2
11
× 72
=
2
= 11 × 36
Sn
= 396
 Sum of first 11 positive integers which are divisible by 6 is 396.
EXERCISE - 1.4 (TEXT BOOK PAGE NO. 14) :
3.
Sol.
Find S10 if a = 6 and d = 3.
For an A.P. a = 6, d = 3
n
Sn =
[2a + (n – 1) d]
2
10
 S 10 =
[2a + (10 – 1) 1]
2
 S 10 = 5 [2 (6) + 9 (3)]
 S 10 = 5 (12 + 27)
 S 10 = 5 (39)
(2 marks)
 S 10 = 195
EXERCISE - 1.4 (TEXT BOOK PAGE NO. 14) :
4.
Sol.
16
Find the sum of all numbers from 1 to 140 which are divisible by 4.
(4 marks)
The natural numbers from 1 to 140 that are divisible by 4 are as follows :
4, 8, 12, 16, .............., 140
S C H O O L S E C TI O N
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











ALGEBRA
These numbers from an A.P. with a = 4, d = t2 – t1 = 8 – 4 = 4
Let, 140 be the nth term of A.P.
t n = 140
t n = a + (n – 1) d
140 = 4 + (n – 1) 4
140 = 4 + 4n – 4
140 = 4n
140
n =
4
n = 35
140 is 35 term of A.P.
We have to find sum of 35 terms i.e. S35,
n
Sn =
[2a + (n – 1)d]
2
35
S 35 =
[2 (4) + (35 – 1) 4]
2
35
S 35 =
[8 + 34 (4)]
2
35
S 35 =
[8 + 136)
2
35
S 35 =
[144]
2
S 35 = 2520
 Sum of natural numbers from 1 to 140 that are divisible by 4 is 2520.
EXERCISE - 1.4 (TEXT BOOK PAGE NO. 14) :
5.
Sol.
Find the sum of the first n odd natural numbers.
Hence find 1 + 3 + 5 + ... + 101.
The first n odd natural numbers are as follows :
1, 3, 5, 7, ............., n
a = 1, d = t2 – t1 = 3 – 1 = 2
n
[2a + (n – 1)d]
Sn =
2
n
 Sn =
[2 (1) + (n – 1) 2]
2
n
 Sn =
[2 + 2n – 2]
2
n
=
[2n]
2
 Sn = n 2
......(i)
1 + 3 + 5 + ........ + 101
Let, 101 be the nth term of A.P.
t n = 101
t n = a + (n – 1) d
 101 = a + (n – 1) d
 101 = 1 + (n – 1) 2
 101 = 1 + 2n – 2
 101 = 2n – 1
 101 + 1 = 2n
 2n = 102
 n = 51
S C H O O L S E C TI O N
(4 marks)
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 101 is the 51st term of A.P.,
 We have to find sum of 51 terms i.e. S51,
Sn = n 2
[From (i)]
 S 51 = (51)2
 S 51 = 2601
EXERCISE - 1.4 (TEXT BOOK PAGE NO. 14) :
6.
Sol.
Obtain the sum of the 56 terms of an A. P. whose 19th and 38th terms are
(4 marks)
52 and 148 respectively.
t19 = 52, t38 = 148
= a + (n – 1) d
tn
 t 19
= a + (19 – 1) d
 52
= a + 18d
 a + 18d= 52
......(i)
t 38
= a + (38 – 1) d
 148
= a + 37d
 a + 37d= 148
......(ii)
Adding (i) and (ii) we get,
a + 18d = 52
a + 37d = 148
2a + 55d = 200
n
=
[2a + (n – 1)d]
Sn
2
56
 S 56
=
[2a + (56 – 1) d]
2
56
 S 56
=
[2a + 55d]
2
56
 S 56
=
[200]
2
 S 56
= 5600
 Sum of first 56 terms of A.P. is 5600.
EXERCISE - 1.4 (TEXT BOOK PAGE NO. 14) :
7.
Sol.
The sum of the first 55 terms of an A. P. is 3300. Find the 28th term.
(3 marks)
S 55
= 3300
[Given]
Sn
 S 55
 3300
 3300
n
[2a + (n – 1)d]
2
55
=
[2a + (55 – 1) d]
2
55
=
[2a + 54d]
2
55
 2 [a + 27d]
=
2
=
3300
= a + 27d
55
300

= a + 27d
5
 a + 27d = 60

18
......(i)
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tn
 t 28
 t 28
 t 28
=
=
=
=
a + (n – 1) d
a + (28 – 1) d
a + 27d
60
[From (i)]
 Twenty eighth term of A.P. is 60.
EXERCISE - 1.4 (TEXT BOOK PAGE NO. 14) :
8.
Sol.
Find the sum of the first n even natural numbers. Hence find the sum
(4 marks)
of first 20 even natural numbers.
The first n even natural numbers are as follows
2, 4, 6, 8, ...........
These numbers form an A.P. with a = 2, d = t2 – t1 = 4 – 2 = 2
n
=
[2a + (n – 1)d]
Sn
2
n
 Sn
=
[2 (2) + (n – 1) 2]
2
n
 Sn
=
[4 + 2n – 2]
2
n
 Sn
=
[2n + 2]
2
n
 Sn
=
× 2 (n + 1)
2
 Sn
= n (n + 1)
 S 20
= 20 (20 + 1)
 S 20
= 20 (21)
 S 20
= 420
 Sum of first twenty even natural numbers is 420.
PROBLEM SET - 1 (TEXT BOOK PAGE NO. 162) :
8.
Sol.
For an A.P. given below find t20 and S10.
For the A.P
a
=
 d
=
 d
=
 d
=
tn
=
t 20
=
1 1 1
, , , ......
6 4 3
(4 marks)
1 1 1
, , , ......
6 4 3
1
6
1 1
–
4 6
3
2
–
12 12
1
12
a + (n – 1)d
1
 1 
 (20 – 1)  
6
 12 
1 19

6 12
2
19

=
12 12
=
S C H O O L S E C TI O N
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21
12
7
 t 20
=
4
n
Now, Sn =
[2a + (n – 1)d]
2
10
[2a  (n – 1)d]
 S 10 =
2
=
 1
 1 
= 5 2    9   
 12  
 6
1 3 
= 5  
3 4
4  9
= 5

 12 
13 
= 5 
12 
 S 10
=
65
12
PROBLEM SET - 1 (TEXT BOOK PAGE NO. 162) :
5.
Sol.
From an A.P. first and last term is 13 and 216 respectively. Common
difference is 7. How many terms are there in that A.P. Find the sum of
all terms.
(4 marks)
for an A.P a = 13
Let last term be tn = 216
d
= 7
tn
= a + (n – 1)d
 tn
= 13 + (n – 1)7
 216 = 13 + 7n – 7
 216 = 6 + 7n
 216 – 6 =7n
 210 = 7n
 n
= 30
n
=
[2a + (n – 1)d]
Sn
2
30
 S 30 =
[2a + (30 – 1)d]
2
 S 30 = 15 [2(13) + 29 (7)]
 S 30 = 15 [26 + 203]
= 15 (229)
 S 30 = 3435

Sum of all terms of AP is 3435.
PROBLEM SET - 1 (TEXT BOOK PAGE NO. 162) :
6.
Sol.
20
Second and fourth term of on A.P. is 12 and 20 respectively. Find the
sum of first 25 terms of that A.P.
(4 marks)
t2 = 12, t4 = 20
tn
= a + (n – 1)d
S C H O O L S E C TI O N
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ALGEBRA
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




t2
= a + (2 – 1)d
12
= a+d
a +d = 12
t4
= a + (4 – 1)d
20
= a + 3d
a + 3d= 20
Subtracting (ii) from (i),
a + d = 12
a + 3d = 20
(–) (–)
(–)
–2d = –8
d = 4
Substituting d = 4 in (i),
a + 4 = 12
a
= 12 – 4
a
= 8
Sn
 S 25
=
n
[2a + (n – 1)d]
2
=
25
[2a + (25 – 1)d]
2
=
25
[2(8) + 24 (4)]
2
=
25
[16 + 96]
2
.......(i)
.......(ii)
25
[112]
2
= 1400
=
S 25
 Sum of 25 terms of the A.P is 1400.
PROBLEM SET - 1 (TEXT BOOK PAGE NO. 162) :
4.
Sol.
In the A.P. 7, 14, 21, ....... How many terms are have to consider for
getting sum 5740.
(4 marks)
for the A.P. 7, 14, 21
a = 7, d = 7
S n = 5740
n
[2a + (n – 1) d]
Sn =
2
n
[ 2(7) + (n – 1)7]
Sn =
2
n
[ 14 + 7n – 7]
5740 =
2
n
5740 =
[ 7n + 7]
2
11480 = 7n2 + 7n
2
 7n + 7n – 11480 = 0
Dividing through at by 7 we get,
n2 + n – 1640 = 0
2
 n + 41n – 40n – 1640 = 0
 n (n + 41) – 40 (n + 41) = 0

(n + 41) (n – 40) = 0
S C H O O L S E C TI O N
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 n + 41 = 0 or
n – 40 = 0
 n = – 41
or
n = 40
n = – 41 is not acceptable because no of terms cannot be negative
 n = 40
 For the given sequence 40 terms have to be considered for getting
sum of 5740.
o
Properties of an A.P. :
PROPERTY - I
For an A.P. with first term ‘a’ and the common difference ‘d’, if any real number
‘k’ is added to each term of an A.P. then the new sequence is also an A.P. with
the first term. ‘a + k’ and the same common difference ‘d’.
For example : For an A.P. 5, 8, 11, 14, ......
a = 5, d = 3
Add any real number k to each term of the A.P. we get
5 + k, 8 + k, 11 + k, 14 + k ......
t1 = 5 + k
t2 – t1 = 8 + k – (5 + k) = 8 + k – 5 – k = 3
t3 – t2 = 11 + k – (8 + k) = 11 + k – 8 – k = 3
t4 – t3 = 14 + k – (11 + k) = 14 + k – 11 – k = 3
PROPERTY - II
For an A.P. with the first term ‘a’ and the common difference ‘d’, if each term
of an A.P. is multiplied by any real number k, then the new sequence is also
an A.P. with the first term ‘ak’ and the common difference ‘dk’.
For example : For an A.P. 5, 8, 11, 14, ......
Multiplying all the terms by any real number k we get,
5k, 8k, 11k, 14k ......
t1 = 5k
t2 – t1 = 8k – 5k = 3k
t3 – t2 = 11k – 8k = 3k
t4 – t3 = 14k – 11k = 3k
(i)
(ii)
(iii)
We assume three, four or five consecutive terms in an A.P. in the
following manner :
Three consecutive terms as a – d, a, a + d
Four consecutive terms as a – 3d, a – d, a + d, a + 3d
(here note that the common difference is 2d)
Five consecutive terms as a – 2d, a – d, a, a + d, a + 2d
EXERCISE - 1.5 (TEXT BOOK PAGE NO. 18) :
1.
Sol.
22
Find four consecutive terms in an A.P. whose sum is 12 and the sum of
3rd and 4th term is 14.
(4 marks)
Let the four consecutive terms in an A.P. be
a – 3d, a – d, a + d, a + 3d
As per the first condition,
a – 3d + a – d + a + d + a + 3d = 12

4a = 12

a = 3
As per the second condition,
a + d + a + 3d = 14

2a + 4d = 14

2 (3) + 4d = 14
S C H O O L S E C TI O N
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



 a
a
a
a
6 + 4d = 14
4d = 14 – 6
4d = 8
d = 2
– 3d = 3 – 3 (2) = 3 – 6 = – 3
–d=3–2=1
+d=3+2=5
+ 3d = 3 + 3 (2) = 9
 The four consecutive terms of A.P. are – 3, 1, 5 and 9.
EXERCISE - 1.5 (TEXT BOOK PAGE NO. 18) :
2.
Sol.
Find four consecutive terms in an A.P. whose sum is –54 and the sum of
1st and 3rd term is – 30.
(4 marks)
Let the four consecutive terms is an A.P. be a – 3d, a – d, a + d and a + 3d
As per first condition,
a – 3d + a – d + a + d + a + 3d = – 54

4a = – 54
–54

a =
4
27

a = –
2
As per the second condition,
a – 3d + a + d = – 30

2a – 2d = – 30
 27 
 2 –
 – 2d
 2 

– 27 – 2d

– 2d

– 2d

 a – 3d =
 a–d
=
 a+d
=
 a + 3d =
= – 30
= – 30
= – 30 +
= –3
3
d =
2
3
27
–
–3  
2
2
27 3
–
–
2
2
27 3
–

2
2
27
3
–
3 
2
2
27
=
=
=
=
– 27 – 9
2
– 30
2
27 3
–

2
2
– 27  9
2
=
– 36
= – 18
2
= – 15
– 24
= – 12
2
– 27  9
–18
=
=
=–9
2
2
=
 The four consecutive terms of an A.P. are – 18, – 15, – 12 and – 9.
EXERCISE - 1.5 (TEXT BOOK PAGE NO. 18) :
3.
Sol.
Find three consecutive terms in an A.P. whose sum is – 3 and the product
of their cubes is 512.
(4 marks)
Let three consecutive terms in an A.P. be
a – d, a, a + d
As per the first given condition,
a–d+a+a+d=–3

3a = – 3

a = –1
S C H O O L S E C TI O N
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











As per the second given condition,
(a – d)3 a3 (a + d)3 = 512
[(a – d) a (a + d)]3 = 512
Taking cube roots on both sides
(a – d) a (a + d) = 3 512
(a – d) a (a + d) = 8
a (a – d) (a + d) = 8
a (a2 – d2) = 8
– 1 [(– 1)2 – d2] = 8
– 1 (1 – d2) = 8
d2 – 1 = 8
d2 = 8 + 1
d2 = 9
d = +3
If d = 3
a–d=–1–3=–4
a–d
a+d=–1+3=2
a+d
=
=
–1–(–3)
–1– 3
=
=
EDUCARE LTD.
2
4
 The three consecutive terms of A.P. are – 4, – 1, 2 or 2, –1, –4
EXERCISE - 1.5 (TEXT BOOK PAGE NO. 18) :
4.
Sol.
In winter, the temperature at a hill station from Monday to Friday is in
A.P. The sum of the temperatures of Monday, Tuesday and Wednesday
is zero and the sum of the temperatures of Thursday and Friday is 15.
(4 marks)
Find the temperature of each of the five days.
Let the temperatures of hill station from Monday to Friday which form
are A.P. be
a – 2d, a – d, a, a + d, a + 2d respectively.
As per the first condition,
a – 2d + a – d + a = 0

3a – 3d = 0

3 (a – d) = 0

a–d = 0

a = d
As per the second condition,
a + d + a + 2d = 15

2a + 3d = 15

2a + 3a = 15
[ a = d]

5a = 15

a = 3

d = 3
[ d = a]
 a – 2d = 3 – 2 (3) = 3 – 6 = – 3
 a–d=3–3=0
 a+d=3+3=6
 a + 2d = 3 + 2 (3) = 3 + 6 = 9
 The temperatures from Monday to Friday are – 3, 0, 3, 6 and 9 respectively.
o
WORD PROBLEMS FOR A.P.
EXERCISE - 1.6 (TEXT BOOK PAGE NO. 20) :
1.
Sol.
24
Mary got a job with a starting salary of Rs. 15000/- per month. She will
get an incentive of Rs. 100/- per month. What will be her salary after
20 months?
(4 marks)
Since Mary’s salary increases by Rs. 100 every month the successive
salaries are in A.P.
S C H O O L S E C TI O N
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EDUCARE LTD.





ALGEBRA
Starting salary of Marry (a) = Rs. 15000
Monthly incentive in salary (d) = 100
No. of months (n) = 20
Salary after twenty months = t20 = ?
t n = a + (n – 1) d
t 20 = a + (20 – 1) d
t 20 = 15000 + 19 (100)
t 20 = 15000 + 1900
t 20 = 16900
 Marry salary after twenty months is Rs. 16900.
EXERCISE - 1.6 (TEXT BOOK PAGE NO. 20) :
2.
Sol.
The taxi fare is Rs. 14 for the first kilometer and Rs. 2 for each additional
kilometer. What will be fare for 10 kilometers ?
(3 marks)
Since the taxi fare increases by Rs. 2 every kilometer after the
first, the successive taxi fares form an A.P.
The taxi fare for first kilometer (a) = Rs. 14
Increase in taxi fare in every kilometer after first kilometer (d) = 2
No. of kilometers covered by taxi (n) = 10
Taxi fare for 10 kilometers = t10 = ?
t n = a + (n + 1) d
 t 10 = a + (10 – 1) d
 t 10 = 14 + 9 (2)
 t 10 = 14 + 18
 t 10 = 32
 Taxi fare for ten kilometers is Rs. 32.
EXERCISE - 1.6 (TEXT BOOK PAGE NO. 20) :
3.
Sol.
Mangala started doing physical exercise 10 minutes for the first day.
She will increase the time of exercise by 5 minutes per day, till she
reaches 45 minutes. How many days are required to reach 45 minutes ?
(3 marks)
Since the workout time Mangala increases by 5 minutes everyday
after the first day, the successive workout times are in A.P.
Workout time for first day (a) = 10 minutes.
Increases in workout time (d) = 5 minutes
Let No. of days required to reach workout time of 45 minutes be ‘n’ days.
t n = 45
 t n = a + (n – 1) d
 45 = 10 + (n – 1) 5
 45 = 10 + 5n – 5
 45 = 5 + 5n
 45 – 5 = 5n
 5n = 40
 n = 8
 8 days required to reach work out time of 45 minutes.
EXERCISE - 1.6 (TEXT BOOK PAGE NO. 20) :
4.
Sol.
There is an auditorium with 35 rows of seats. There are 20 seats in the
first row, 22 seats in the second row, 24 seats in the third row, and so
on. Find the number of seats in the twenty fifth row.
(3 marks)
Since the no. of seats in each row of the auditorium are 20, 22, 24, ......
The no. of seats in each row form an A.P.
S C H O O L S E C TI O N
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



EDUCARE LTD.
No. of seats in first row (a) = 20
Difference in no. of seats in two successive rows is (d) = 2
No. of seats in 25th row = t25 = ?
t n = a + (n + 1) d
t 25 = a + (25 – 1) d
t 25 = 20 + 24 (2)
t 25 = 20 + 48
t 25 = 68
 There are 68 seats in 25th row.
EXERCISE - 1.6 (TEXT BOOK PAGE NO. 20) :
5.
Sol.
A village has 4000 literate people in the year 2010 and this number
increases by 400 per year. How many literate people will be there till the
year 2020 ? Find a formula to know the number of literate people after
n years ?
(4 marks)
Since the no. of literate people increases by 400 every year, the
population of literate people every year forms an A.P.
No. of people in the year 2010 (a) = 4000
Increase in population every year (d) = 400
No. of years from 2010-2020 (n) = 10
t n = a + (n – 1) d
 t 10 = 4000 + (10 – 1) 400
 t 10 = 4000 + 9 (400)
 t 10 = 4000 + 3600
 t 10 = 7600
 There will be 7600 literate people till the year 2020
t n = a + (n – 1) d
 t n = 4000 + (n – 1) 400
 t n = 4000 + 400n – 400
 t n = 3600 + 400n
There will be (3600 + 400n) literate people after n years.
EXERCISE - 1.6 (TEXT BOOK PAGE NO. 20) :
6.
Sol.
Neela saves in a ‘Mahila Bachat gat’ Rs. 2 on the first day, Rs.4 on the
second day, Rs. 6 on the third day and so on. What will be her saving in
the month of February 2010 ?
(4 marks)
The savings done by Neela on each day is as follows 2, 4, 6, .......
These every day savings form an A.P. with
First day saving (a) = 2
Difference in savings made in two successive days (d) = 2
Total no. of days in the month of February 2010 (n) = 28
 Total savings for the month of February (S28) = ?
n
[2a + (n – 1) d]
Sn =
2
28
 S 28 =
[2 (2) + (28 – 1) (2)]
2
= 14 [4 + 27 (2)]
= 14 [4 + 54]
 S 28 = 14 [58]
 S 28 = 812
 Neela saved Rs. 812 in the month of February.
26
S C H O O L S E C TI O N
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EXERCISE - 1.6 (TEXT BOOK PAGE NO. 20) :
7.
Sol.
Babubhai borrows Rs. 4000 and agrees to repay with a total interest of
Rs. 500. in 10 instalments, each instalment being less that the preceding
instalment by Rs. 10. What should be the first and the last instalment?
(4 marks)
Total money repaid by Babubhai in 10 instalments = (S 10 )
= 4000 + 500
= Rs. 4500
No. of instalments (n) = 10
Difference between two consecutive instalments (d) = – 10
First instalment = (a) = ?
Last instalment (t10) = ?
n
[2a + (n – 1) d]
Sn =
2
10

S 10 =
[2a + (10 – 1) d]
2

4500 = 5 [2a + 9 (– 10)]









4500
=
5
900 =
900 + 90 =
990 =
990
=
2
a =
tn =
t 10 =
t 10 =
t 10 =
t 10 =
2a – 90
2a – 90
2a
2a
a
495
a + (n – 1) d
a + (10 – 1) d
495 + 9 (– 10)
495 – 90
405
 First instalment is Rs. 495 and last instalment is Rs.405.
EXERCISE - 1.6 (TEXT BOOK PAGE NO. 20) :
8.
Sol.
A meeting hall has 20 seats in the first row, 24 seats in the second row,
28 seats in the third row, and so on and has in all 30 rows. How many
seats are there in the meeting hall ?
(4 marks)
The no. of seats in each row are as follows 20, 24, 28, .........
The no. of seats in each row form an A.P. with
First term (a) = 20
Difference between the no. of seats in two successive rows (d) = 4
Total no. of rows (n) = 30
Total no. of seats in 30 rows (S30) = ?
Sn =
n
[2a + (n – 1) d]
2
30
[2 (20) + (30 – 1) 4]
2
= 15 [40 + 116]
= 15 (156)
= 2340

S 30 =



S 30
S 30
S 30
 Total no. of seats in the meeting hall is 2340.
S C H O O L S E C TI O N
27
ALGEBRA
MT
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EXERCISE - 1.6 (TEXT BOOK PAGE NO. 20) :
9.
Sol.
Vijay invests some amount in National saving certificate. For the 1st year
he invests Rs. 500, for the 2nd year he invests Rs. 700, for the 3rd year
he invests Rs. 900, and so on. How much amount he has invested in
12 years ?
(4 marks)
Yearly investments of Vijay are as follows 500, 700, 900, .......
The yearly investments form an A.P. with first year investment (a) = 500
Difference between investment done in two successive years (d) = 200.
No. of years (n) = 12
Total investment done in 12 years = (S12) = ?
n
[2a + (n – 1) d]
Sn =
2
12
 S 12 =
[2 (500) + (12 – 1) 200]
2
 S 12 = 6 [1000 + 11 (200)]
 S 12 = 6 [1000 + 2200]
 S 12 = 6 [3200]
 S 12 = 19200
 Total investment done in 12 years is Rs. 19200.
EXERCISE - 1.6 (TEXT BOOK PAGE NO. 20) :
10.
Sol.
In a school, a plantation program was arranged on the occasion
of world environment day, on a ground of triangular shape.
The trees are to be planted as shown in the figure. One plant
in the first row, two in the second row, three in the third row
and so on. If there are 25 rows then find the total number of
(4 marks)
plants to be planted.
The number of trees in each row upto the 25th row areas follows :
1, 2, 3, 4, ...........
These no. of trees planted in each row forms an A.P. with
No. of trees in first row (a) = 1
Difference between no. of trees planted in two successive rows (d) = 1
No. of rows (n) = 25
n
Sn =
[2a + (n – 1) d]
2
25
 S 25 =
[2 (1) + (25 – 1) 1]
2
25
 S 25 =
[2 + 24]
2
25
 S 25 =
(26)
2
 S 25 = 25 (13)
 S 25 = 325
•
• •
•• •
•• ••
 325 trees were planted in 25 rows.
o
GEOMETRIC PROGRESSION :
Geometric progression is a sequence such that the given first term each term
so obtained by multiplying a non-zero constant ‘r’ to the preceeding term.
Also it is a sequence in which the ratio of to two consecutive terms in the
progression is constant.
Consider, 2, 4, 8, 16, ......
t2
t3
t 4 16
4
8
We have t  2 = 2, t  4  2 , t  8  2 and so on.
3
1
2
28
S C H O O L S E C TI O N
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ALGEBRA
Here for a G.P. first term is denoted as ‘a’ and common ratio is denoted as ‘r’.
General term of a G.P.
Thus in general any G. P. can be expressed as a, ar, ar2, ar3, ‘a’ is first
term and r is the common ratio.
The nth term of a G.P.
Consider a G.P. whose first term is ‘a’ and the common ratio is ‘r’
t1, t2, t3, t4, .......
Then, t1 = a, t2 = ar, t3 = ar2, t4 = ar3
t2
t3
tn
t1 = r, t2 = r, ......., t n–1 = r
Multiplying all (n – 1) ratios we get,
t2
t
t
t
 3  4  ......  n = r × r × r × r × r × r ..... (n – 1) times
t1
t2
t3
t n– 1
tn
tn
(n–1)
i.e.
= r(n–1), tn = ar(n–1)
t1 = r
a
Hence in general tn = arn–1
Thus the nth term of a G.P. with the first term ‘a’ and the common ratio ‘r’ is
 tn = arn–1
EXERCISE - 1.7 (TEXT BOOK PAGE NO. 25) :
1.
Sol.
Find the ninth term of the G.P. 3, 6, 12, 24, ....
For the G.P.
3, 6, 12, 24, ........
a=3
t2
6
r= t =
=2
3
1
n – 1
t n = ar
 t 9 = 3 (2)9 – 1
 t 9 = 3 (2)8
 t 9 = 3 (256)
 t 9 = 768
(3 marks)
 The ninth term of the G.P. is 768.
EXERCISE - 1.7 (TEXT BOOK PAGE NO. 25) :
2.
Sol.
Write down the first five terms of the geometric progression which has
(2 marks)
first term 1 and common ratio 4.
For the G.P.
The first term (a) = 1
Common ratio (r) = 4
= 1
t1 = a
t 2 = ar = 1 × 4
= 4
t 3 = ar 2 = 1 × (4)2 = 16
t 4 = ar 3 = 1 × (4)3 = 64
t 5 = ar 4 = 1 × (4)4 = 256
 The first five terms of the G.P. are 1, 4, 16, 64 and 256.
EXERCISE - 1.7 (TEXT BOOK PAGE NO. 25) :
3.
Sol.
Find the 4th and 9th terms of the G.P. with first term 4 and common ratio
2.
(3 marks)
For the G.P.
The first term (a) = 4
Common ratio (r) = 2
S C H O O L S E C TI O N
29
MT
ALGEBRA








tn
t4
t4
t4
t4
t4
t9
t9
t9
t9
=
=
=
=
=
=
=
=
=
=
EDUCARE LTD.
arn – 1
ar4 – 1
ar 3
4 (2)3
4 (8)
32
ar9 – 1
ar 8
4 (2)8
1024
 The fourth and ninth term of the G.P. are 32 and 1024 respectively.
EXERCISE - 1.7 (TEXT BOOK PAGE NO. 25) :
4.
Sol.
Find the common ratio and the 7th term of the G.P. 2, – 6, 18, .... (2 marks)
For the G.P. 2, – 6, 18, ........
The first term (a) = 2
6
Common ratio (r) = – = – 3
2
Seventh term t7 = ?
t n = arn – 1
 t 7 = ar7 – 1
 t 7 = ar 6
 t 7 = 3 (–3)6
 t 7 = 2 (729)
 t 7 = 1458
 The common ratio of the G.P. is – 3 and the seventh term is 1458.
EXERCISE - 1.7 (TEXT BOOK PAGE NO. 25) :
5.
Sol.
Find the 69th term of the G.P. 1, –1, 1, –1, ....
For the G.P. 1, – 1, 1, – 1
First term (a) = 1
–1
=–1
Common ratio (r) =
1
n – 1
t n = ar
 t 69 = ar69 – 1
 t 69 = ar 68
 t 69 = 1 (– 1)68
 t 69 = 1 × 1
 t 69 = 1
(2 marks)
 Sixtyninth term of G.P. is 1.
EXERCISE - 1.7 (TEXT BOOK PAGE NO. 25) :
6.
Sol.
Find the 15th term of the G.P. 3, 12, 48, 192, ...
For the G.P. 3, 12, 48, 192, ........
First term (a) = 3
12
= 4
Common ratio (r) =
3
n – 1
t n = ar
 t 15 = ar15 – 1
 t 15 = ar 14
 t 15 = 3 (4)14
 t 15 = 3 × 414
(2 marks)
 Fifteenth term of G.P. is 3 × 414.
30
S C H O O L S E C TI O N
MT
o
ALGEBRA
EDUCARE LTD.
Sum of first n terms of a G.P. :
Suppose that we want to find the sum of the first n terms of a geometric
progression a, ar, ar2, ar3 ...... where r  1.
Let, Sn = a + ar + ar2 + ar3 + ..... + ar(n–1)
......(i)
Multiplying the above equation by r we get,
rSn = ar + ar2 + ar3 + ..... + ar(n–1) + arn
.....(ii)
Subtracting equation (ii) from equation (i) we get,
Sn – rSn = a – arn
So that, Sn (1 – r) = a (1 – rn)
Dividing by (1 – r) (since r  1) we get,
a (1 – r n )
Sn =
[We use this if r < 1]
1– r
This can also be written as
a (r n – 1)
[We use this if r > 1]
Sn =
r –1
NOTE : When the common ratio r = 1, the G.P. becomes a, a, a, a, .....
In this case clearly, the sum of the n terms is a + a + a + .....,
n times = na.
EXERCISE - 1.8 (TEXT BOOK PAGE NO. 27) :
1.
(i)
Sol.
Find the indicated sums for the following Geometric Progressions.
2, 6, 18, ... Find S7.
(2 marks)
a (1 – r n )
2, 6, 18, .........
Sn =
1– r
a
= 2
a (1 – r 9 )
6

S9 =
1– r
r
=
2
2 (1 – (– 2)9 )
= 3

S9 =
1 – (– 2)
a (1 – r n )
2 (1 – (– 512))
Sn =
1– r

S9 =
1 2
7
a (1 – r )
2 (1  512)
 S7 =

S9 =
1– r
3
 S7
2 (1 – 37 )
=
1–3
 S7 =
 S7 =
 S7 =
 S7 =
2 (1 – 37 )
–2
– 1 (1 – 37)
37 – 1
2187 – 1

S9 =
1026
3

S9 =
342
S 12 =
a (1 – r12 )
1– r

S 12 =
2 (1 – (– 2)12 )
1 – (– 2)

S 12 =
2 (1 – 4096)
3

S 12 =

S 12 =
2  (– 4095)
3
2 (– 1365)

S 12 =
– 2730
 S 7 = 2186
(ii)
Sol.
2, – 4, 8, –16, ... Find S9 and S12.
(3 marks)
For the G.P. 2, – 4, 8, – 16, .......
a = 2
–4
r =
2
= –2
S C H O O L S E C TI O N
31
MT
ALGEBRA
(iii)
1,
1
2
,
1
2
2
1
,
3
2
, ... Find S6.
(iv)
Sol.
(2 marks)
Sol.
a
r
= 1
=
1
2
1
1
=
2
Sn
a (1 – r )
=
1– r
 S6 =

1 1 –



2 , 2, ... Find S10.
For the G.P.
a = 5
r =
2
1 
  
2 
1
1–
2
S 10 =
6
1
64
1
2
1–
 S6 =
1,
Sn =
n

EDUCARE LTD.
S 10 =
a (1 – r n )
1– r

a 1 – r10
(3 marks)

1– r
 
1 1 – 2

1– 2
10



10

S 10 =

S 10 =
 S6 =
64 – 1 1

64
2

S 10 =
 S6 =
63 2

64 1

S 10 =
 S6 =
63
32

S 10 =

S 10 =

S 10 =

S 10 =

S 10 =
 1
1 –  22 
 
 
1– 2
1 – 25
1– 2
1 – 32
1– 2
– 31
1– 2
 2
1 – 2  1  2 
– 31 1  2 
1 –  2
– 31 1  2 
– 31 1 
2
2
1–2

– 31 1 
2

–1

31 1 
2

EXERCISE - 1.8 (TEXT BOOK PAGE NO. 27) :
2.
Sol.
32
If in a G.P. r = 2 and t8 = 64 then find a and S6.
For a G.P.
r
= 2
t 8 = 64
t n = arn – 1
t 8 = ar8 – 1
64 = ar 7
64 = a (2)7
64 = a (128)
64
= a
128
(3 marks)
S C H O O L S E C TI O N
MT
ALGEBRA
EDUCARE LTD.
 a
Sn
1
2
a (1 – r n )
=
1– r
=
 S6 =






a 1 – r6

1– r
1
(1 – 26 )
2
S6 =
1–2
1
(1 – 64)
S6 = 2
–1
1
(– 63)
S6 = 2
–1
63
S6 =
2
1
63
a=
and S6 =
.
2
2
EXERCISE - 1.8 (TEXT BOOK PAGE NO. 27) :
3.
Sol.
If S3 = 31 and S6 = 3906 then find a and r.
For a G.P.
a (1 – r n )
Sn =
1– r
a (1 – r 3 )
S3 =
1– r
But,
S 3 = 31
[Given]

a (1 – r 3 )
= 31
1– r
Similarly,
a (1 – r 6 )
S6 =
1– r
But,
S 6 = 3906
(4 marks)
.......(i)
[Given]
6


a (1 – r )
= 3906
1– r
Dividing (ii) by (i),
a (1 – r 6 )
1– r

1–r
a (1 – r 3 )
1 – r6
1 – r3


S C H O O L S E C TI O N
......(ii)
=
3906
31
= 126
1 – (r 3 )2
= 126
1 – r3
(1  r 3 ) (1 – r 3 )
= 126
1 – r3
33
MT
ALGEBRA









EDUCARE LTD.
1 + r3 = 126
r 3 = 126 – 1
r 3 = 125
Taking cube roots on both sides,
r = 5
Substituting r = 5 in (i),
a (1 – 53 )
= 31
1–5
a (1 – 125)
= 31
–4
a (– 124) = 31 × – 4
– 124a = – 124
–124
a = –124
a = 1
a = 1 and r = 5.
EXERCISE - 1.8 (TEXT BOOK PAGE NO. 27) :
4.
Sol.
If S6 = 126 and S3 = 14 then find a and r.
For a G.P.
Sn
a (1 – r n )
=
1– r
a (1 – r 6 )
 S6 =
1– r
But, S 6 = 126

(4 marks)
[Given]
a (1 – r 6 )
= 126
1– r
Similarly,
a (1 – r 3 )
S3 =
1– r
But, S 3 = 14
.....(i)
[Given]
3








34
a (1 – r )
= 14
1– r
Dividing (i) by (ii),
 a (1 – r 6 )   a (1 – r 3 ) 


 =
 1– r   1– r 
a (1 – r 6 )
1– r

=
1–r
a (1 – r 3 )
1 – r6
=
1 – r3
2
3 2
1 – (r )
=
1 – r3
(1  r 3 ) (1 – r 3 )
=
1 – r3
1 + r3 =
r3 =
r3 =
Taking cube roots on both
r =
.....(ii)
126
14
9
9
9
9
9
9–1
8
sides
2
S C H O O L S E C TI O N
MT
EDUCARE LTD.





ALGEBRA
Substituting r = 2 in (ii), we get,
a (1 – 23 )
= 14
1–2
a (1 – 8)
= 14
–1
a (– 7)
= 14
–1
7a = 14
a = 2
a = 2, r = 2.
EXERCISE - 1.8 (TEXT BOOK PAGE NO. 27) :
5.
Sol.
If the nth, (2n)th, (3n)th terms of a G.P. are a, b, c respectively then show
(4 marks)
that b2 = ac.
For the G.P.
Let first term be A
Common ratio be r

t n = Arn – 1
But,t n = a
[Given]
n – 1
 Ar
= a
......(i)
t 2n = Ar2n – 1
But,
t 2n = b
[Given]
 Ar2n – 1 = b
......(ii)
t 3n = Ar3 – 1
But,
[Given]
t 3n = c
 Ar3n – 1 = c
.....(iii)
b = Ar2n – 1
[From (ii)]
Squaring both sides,
b2 = (Ar2n – 1)2

b2 = A2 (r2n – 1)2

b2 = A2 r4n – 2
......(iv)
Multiplying (i) and (iii) we get,
ac = Arn – 1 . Ar3n – 1

ac = A2 rn – 1 + 3n – 1

ac = A2 r4n – 2
.....(v)
 From (iv) and (v) we get,

b 2 = ac
PROBLEM SET - 1 (TEXT BOOK PAGE NO. 163) :
9.
Sol.
In a school, tree plantation on Independence day was arranged. Every
student from I standard will plant 2 trees, II standard students will plant
4 trees each, III standard students will plant 8 trees each etc. If there are
5 standard, how many trees are planted by the student of that school?
(4 marks)
No of trees planted by a student from 1st, 2nd, 3rd standards are 2, 4,
8 respectively
The no. of trees planted by each student form a G.P with
4
a=2,r=
=2
2
 There are 5 standards.
 Total no. of trees planted by the student of that school is S5
S C H O O L S E C TI O N
35
MT
ALGEBRA
Sn
a – (1 – r n )
=
1–r
S5
=
S5
=
S5
=
S5
EDUCARE LTD.

2 1 – (x)5

1–2
2 (1 – 32)
1–2
2 (– 31)
–1
= 62
 Total no of trees planted by student of school is 62 trees.
o
ARITHMETIC MEAN :
If three numbers x, y, z are in A.P. then ‘y’ is called the Arithmetic mean
between x and z.
To find the Arithmetic mean between any two numbers x and y.
Suppose x and y are any two numbers. Let the Arithmetic mean between x
and y be ‘A’. Then x, A, y are in A.P.
 A–x=y–A
 2A = x + y
x y
2
Thus the arithmetic mean ‘A’ between any two numbers x and y is given by
 A=
A=
o
x y
2
GEOMETRIC MEAN :
If three numbers x, y, z are in G.P. then ‘y’ is called the geometric mean
between x and z.
The find the geometric mean between any two numbers x and y.
Suppose x and y are any two numbers with the same sign. Let the Geometric
mean between x and y be ‘G’, Then x, G, y are in G.P.
G
y
=
x
G
 G2 = xy

 G =  xy
Thus Geometric mean ‘G’ between any two numbers x and y is given by
G= 
xy
EXERCISE - 1.9 (TEXT BOOK PAGE NO. 31) :
1.
Sol.
36
Find three consecutive terms in a G.P. such that the sum of the first
two terms is 9 and the product of all the three is 216.
(4 marks)
a
Let three consecutive terms of G.P. be
, a, ar
r
As per first given condition,
a
+a=9
......(i)
r
S C H O O L S E C TI O N
MT
ALGEBRA
EDUCARE LTD.
As per second condition,
a
× a × ar = 216
r
a 3 = 216
Taking cube roots on both side,






a = 3 216
a = 6
Substituting a = 6 in (i),
6
+6 = 9
r
6  6r
= 9
r
6 + 6r = 9r
6 = 9r – 6r
3r = 6
r = 2
a
6
=
= 3
r
2
ar = 6 × 2 = 12
 The three consecutive terms of G.P. are 3, 6, 12.

EXERCISE - 1.9 (TEXT BOOK PAGE NO. 31) :
2.
Sol.
Find three consecutive terms in a G.P. such that the sum of the 2nd
and 3rd term is 60 and the product of all the three is 8000. (4 marks)
a
Let three consecutive terms of G.P. be
, a, ar.
r
As per first condition,
a + ar = 60
......(i)
As per second condition,
a
× a × ar = 8000
r
a 3 = 8000
Taking cube roots on both side,
a
=
3
8000
a
=
3
20  20  20

a
Substituting
20 + 20r

20r

20r

20r

r

r
a
20
=
r
2
ar
ar



=
a
=
=
=
=
20
= 20 in (i),
60
60 – 20
60 – 20
40
40
=
20
= 2
= 10
= 20 × 2 .
= 40
 The three consecutive terms of the G.P. are 10, 20, 40.
S C H O O L S E C TI O N
37
ALGEBRA
MT
EDUCARE LTD.
EXERCISE - 1.9 (TEXT BOOK PAGE NO. 31) :
3.
Sol.
38
Sachin, Sehwag and Dhoni together scored 228 runs. Their individual
scores are in G.P. Sehwag and Dhoni together scored 12 runs more than
Sachin. Find their Individual scores.
(5 marks)
a
Let individual scores of Sachin, Sehwag and Dhoni be
, a and ar
r
As per the first given condition,
a
+ a + ar = 228
.....(i)
r
As per second given condition,
a
......(ii)
a + ar = 12 +
r
Substituting (ii) in (i),
a
a
+ 12 +
= 228
r
r
2a

12 +
= 228
r
2a

= 228 – 12
r
2a

= 216
r
a
216

=
r
2
a

= 108
.....(iii)
r

a = 108r
Substituting (iii) in (i),
 a

108r
  108 , a  108r 
+ 108r + (108 r)r = 228

r
r


Multiplying throughout by r,
108 + 108r + 108r2 = 228
2
 108r + 108r + 108 – 228 = 0

108r2 + 108r – 120 = 0
Dividing throughout by 12 we get,
9r2 + 9r – 10 = 0

9r2 + 15r – 6r – 10 = 0

3r (3r – 5) – 2 (3r – 5) = 0

(3r – 5) (3r – 2) = 0
 3r – 5 = 0
or
3r – 2 = 0

3r = 5
or
3r = 2
5
2

r=
or
r=
3
3
5
2
If r =
If r =
3
3
a = 108r
a = 108r
5
2
a = 108 ×
a = 108 ×
3
3
a = 36 × 5
a = 36 × 2
a = 180
a = 72
5
2
 ar = 180 ×
ar = 72 ×
3
3
 ar = 300
ar = 48
S C H O O L S E C TI O N
MT
ALGEBRA
EDUCARE LTD.
5
is not acceptable because ar (individual score) cannot be 300
3
as the sum of the scores is 228.
 ar = 48
 The runs scored by Sachin, Sehwag and Dhoni are 108 runs,
72 runs and 48 runs.
r =
EXERCISE - 1.9 (TEXT BOOK PAGE NO. 31) :
4.
Sol.
If 25 is the arithmetic mean between x and 46, then find x. (2 marks)
 25 is the arithmetic mean between x and 46
x  46
 25 =
2
 50 = x + 46
 x
= 50 – 46
 x
= 4
EXERCISE - 1.9 (TEXT BOOK PAGE NO. 31) :
5.
Sol.
If x + 3 is the geometric mean between x + 1 and x + 6 then find x. (2 marks)
 x + 3 is the geometric mean between x + 1 and x + 6.

(x + 3)2 = (x + 1) (x + 6)
2
 x + 6x + 9 = x2 + 6x + x + 6

9 = x+6

9–6 = x

x = 3
EXERCISE - 1.9 (TEXT BOOK PAGE NO. 31) :
6.
Sol.
Find the geometric mean of
82 – 1 and
Let G be the geometric mean of
 G2 =
 G2 =



82 – 1
82

2
82 + 1 .
82 – 1 and
(2 marks)
82  1

82  1
– 12
 G 2 = 82 – 1
 G 2 = 81
Taking square roots on both sides,
G =
81
 G = +9
 The geometric mean of
81 – 1 and
81  1 is 9 or –9
EXERCISE - 1.9 (TEXT BOOK PAGE NO. 31) :
7.
Sol.
If the arithmetic mean and the geometric mean of two numbers are in
the ratio 5 : 4 and the sum of the two numbers is 30 then find these
numbers.
(4 marks)
Let the two numbers be x and y
Let Arithmetic mean of x and y be denoted as A and geometric
mean of x and y be G.
x y

A =
......(i)
2

G = + xy
.....(ii)
x + y = 30
.....(iii)
[Given]

S C H O O L S E C TI O N
39
MT
ALGEBRA














Substituting (iii) in (i),
30
A =
2
A = 15
A
5
=
G
4
15
5
=
G
4
4
15 
= G
5
G = 12
xy = 12
Squaring both sides,
xy = 144
144
y =
......(iv)
x
Substituting (iv) in (iii), we get,
144
= 30
x+
x
Multiplying throughout by x we get,
x2 + 144 = 30x
2
x – 30x + 144 = 0
x2 – 24x – 6x + 144 = 0
x (x – 24) – 6 (x – 24) = 0
(x – 24) (x – 6) = 0
x – 24 = 0 or x – 6 = 0
x = 24
or x = 6
If x = 24
If x = 6
144
144
y=
y=
x
x
144
144
y=
y=
24
6
y=6
y = 24
EDUCARE LTD.
[Given]
 The two numbers are 6 and 24.
MCQ’s
1.
Select the correct sequence of that numbers satisfies for an G.P.
(a) 1, 3, 6, 10, ....
(b) 3, 6, 12, 24
(c) – 10, – 17, – 16, – 19
(d) 22, 26, 28, 31
2.
Select the correct sequence of that numbers not satisfies for an A.P.
(a) 0.5, 2, 3.5, 5
(b) 22, 26, 28, 31
(c) 3, 5, 7, 9, 11
(d) 1, 4, 7, 10
3.
For an A.P. 4, 9, 14, ............. then t11 = ............. .
(a) 49
(b) 54
(c) 59
(d) 44
4.
The 18th term of an A.P. 1, 7, 13, 19, ............. is
(a) 103
(b) 109
(c) 97
(d) 115
40
S C H O O L S E C TI O N
MT
ALGEBRA
EDUCARE LTD.
5.
The general term of an A.P. is tn ............. .
(a) a – (n – 1) d
(b) a + (n – 1) d
(c) a – (n + 1) d
(d) a + (n + 1) d
6.
Sum of 1st n terms of an A.P. Sn = ............. .
n
n
[a  (n – 1)d]
[2a  (n  1)d]
(a)
(b)
2
2
n
n
[2a  (n – 1)d]
[a  (n  1)d]
(c)
(d)
2
2
7.
If a = 6 and d = 3. S10 = ?
(a) 192
(c) 198
(b)
(d)
195
201
8.
The general form of an G.P. tn = ............. .
a
rn – 1
(b)
(a) n – 1
r
a
(c) arn – 1
(d) ra n– 1
9.
The 9th term of an G.P. 3, 6, 12, 24 is ............. .
(a) 384
(b) 768
(c) 1536
(d) 192
10.
The 7th term of G.P. 2, – 6, 18 is ............. .
(a) – 1458
(b) 1458
(c) 486
(d) – 486
11.
Sum for 1st in terms of G.P. for r > 1 is Sn = ............. .
a (r n – 1)
(a)
r –1
(b)
a (r n – 1 )
1– r
a (r n  1)
r 1
(d)
a (r – 1)
r n –1
(c)
12.
Three consecutive numbers are an G.P. and their product is 1000 then the
second term a is ............. .
1
10
(c) 10
(a)
13.
(b)
1
(d)
100
The 9th term of an G.P. 3, 6, 12, 24, ..... is ............. .
(a) 3(2)8
(b) 3(2)9
 1
(c) 3  – 
 2
8
(d)
1
3 
2
9
14.
If (x + 3) is geometric mean of (x + 1) and (x + 6) then x = ............. .
(a) 9
(b) 6
(c) 15
(d) 3
15.
If 25 is arithmetic mean of x and x + 46 then x = ............. .
(a) 36
(b) 2
(c) 46
(d) 50
S C H O O L S E C TI O N
41
MT
ALGEBRA
EDUCARE LTD.
16.
Arithmetic mean A between any two number is given by A = ............. .
x – y
xy
(b)
(a)
2
2
2
(c) 2 (x + y)
(d) x  y
17.
If G is the geometric mean (GM) of two numbers x, y then ............. .
(a) G2 = xy
(b) G = xy
x y
xy
(d) G 
(c) G 
2
2
18.
Three consecutive numbers are in A.P. such that the sum of the first and
the last is – 8, so that the second term a is ............. .
(a) – 4
(b) – 2
(c) 0
(d) 2
19.
The sum of first 10 natural numbers is ............. .
(a) 55
(b) 155
(c) 310
(d) 210
20.
In G.P. 4 consecutive term is ............. .
a a
a a
2
, , ar, ar 3
(a) 2 , , ar, ar
(b)
r
r
r3 r
a
a
a
a
a
3
(c) – , – ar, +ar,
(d) – 3 , – , + ,  ar
r
r
r
r
r
: ANSWERS :
1.
3.
(b) 3, 6, 12, 24
(b) 54
2.
4.
5.
(b) a + (n – 1) d
6.
7.
9.
(b) 195
(b) 768
8.
10.
(b) 22, 26, 28, 31
(a) 103
n
[2a  (n – 1)d]
(c)
2
n – 1
(c) ar
(b) 1458
12.
(c) 10
13.
a (r n – 1)
r –1
(a) 3 (2)8
14.
15.
(b) 2
16.
(d) 3
x y
(b)
2
(a) – 4
a a
3
(b) 3 , , ar, ar
r
r
11.
(a)
17.
2
(a) G = xy
18.
19.
(a) 55
20.
Open End Questions
1.
Sol.
42
Write first four terms of a G.P. whose first term is 3.
For a G.P.
 a=3
Let us take r = 2
t1 = a
= 3
t 2 = ar
= 3×2
= 6
t 3 = ar 2 = 3 × 2 × 2
= 12
S C H O O L S E C TI O N
MT
t4

2.
Sol.
=
3 × 2 × 2 × 2 = 24
The first four terms of G.P. one of the possibility is 3, 6, 12, 24.
G.P. whose common ratio is – 2.
=
=
=
–4
2×4 = 8
2 (– 8) = – 16
One of the possible ways of writing the first form terms of G.P.
could be 2, – 4, 8 and – 16.
Write first four terms of an A.P. whose first term is 10.
For an A.P. a = 10
Let us take d = 5
= 10
t1 = a
= 15
t 2 = a + d = 10 + 5
t 3 = a + 2d = 10 + 2 (5) = 10 + 10 = 20
t 4 = a + 3d = 10 + 3 (5) = 10 + 15 = 25

4.
Sol.
= ar 3
Write first four terms of a
For a G.P.
 r=–2
Let us take a = 2
= 1
t1 = a
= 2 (– 2)
t 2 = ar
t 3 = ar 2 = 2 (– 2)2
t 4 = ar 3 = 2 (– 2)3

3.
Sol.
ALGEBRA
EDUCARE LTD.
One of the possible ways of writing the first four terms of A.P.
could be 10, 15, 20, 25.
Write first three
For an A.P. d
Let us take a
t1 = a
t2 = a + d
t 3 = a + 2d

terms of A.P. whose common difference in – 3.
=–3
= 5
= 5
= 5 + (– 3) = 5 – 3
= 2
= 5 + 2 (– 3) = 5 – 6
= –1
One of the possible ways of writing first three terms of A.P. could
be 5, 2, – 1.
HOTS PROBLEMS
(Problems for developing Higher Order Thinking Skill)
1.
If the sum of p terms of an A. P. is equal to the sum of q terms then
show that the sum of its p + q terms is zero.
(5 marks)
Given : Sp = S q
Prove : Sp+q = 0
n
Proof :
Sn =
[2a + (n – 1)d]
2
p
[2a + (p – 1) d]
......(i)
Sp =
2
Similarly,
q
[2a + (q – 1) d]
.....(ii)
Sq =
2
Similarly,
pq
[2a + (p + q – 1) d]
.....(iii)
Sp+q =
2
[Given]
 Sp = S q
S C H O O L S E C TI O N
43
MT
ALGEBRA









2.
Sol.
p
q
[2a + (p – 1) d] =
[2a + (q – 1) d]
2
2
Multiplying both sides by 2,
p [2a + (p – 1) d]
p [2a + pd – d]
2ap + p2d – pd
2
2ap + p d – pd – 2aq – q2d + qd
2ap – 2aq + p2d – q2d – pd + qd
2a (p – q) + d (p2 – q2) – d (p – q)
2a (p – q) + d (p + q) (p – q) – d (p – q)
Dividing throughout by p – q we get,
2a + d (p + q) – d = 0
2a + d [p + q – 1] = 0
Substituting (iv) in (iii) we get,
pq
Sp+q =
[0]
2
Sp+q = 0
Hence proved.
=
=
=
=
=
=
=
EDUCARE LTD.
q [2a + (q – 1) d]
q [2a + qd – d]
2aq + q2d – qd
0
0
0
0
......(iv)
How many two digit numbers leave the remainder 1 when divided by 5 ?
(3 marks)
The two digit numbers that leave a remainder of 1 when divided by
5 are as follows
11, 16, 21, 26, 31, ............, 96.
These numbers form an A.P.
With a = 11, d = 5
Let, tn = 96
tn
= a + (n – 1) d

96
= 11 + (n – 1) 5

96
= 11 + 5n – 5

96
= 6 + 5n

96 – 6
= 5n

5n
= 90
90

n
=
5

n
= 18
 There are 18 two digit numbers leaving a remainder of 1 when divided by 5.
3.
Sol.
44
How many terms of the A. P. 16, 14, 12, ..... are needed to given the
sum 60 ? Explain why we get double answer?
(4 marks)
For the A.P. 16, 14, 12, .........
a = 16
d = 14 – 16 = – 2
Let, Sn = 60
n
Sn
=
[2a + (n – 1) d]
2
n

60
=
[2 (16) + (n – 1) (– 2)]
2

120
= n [32 – 2n + 2]

120
= n [34 – 2n]

20
= 34n – 2n2

2n2 – 34n + 120
= 0
Dividing throughout by 2,
S C H O O L S E C TI O N
MT
ALGEBRA
EDUCARE LTD.






n2 – 17n + 60
= 0
n – 12n – 5n + 60
= 0
n (n – 12) – 5 (n – 12) = 0
(n – 12) (n – 5)
= 0
n – 12 = 0
or n – 5 = 0
n = 12
or n = 5
The no. of terms required to give a sum of 60 are 12 or 5.
The reason for getting double answers is the common difference is
– 2 and the progression is towards the negative side.
The A.P. is as follows :
16, 14, 12, 10, 8, 6, 4, 2, 0, – 2, – 4, – 6 ....
S5 = 16 + 14 + 12 + 10 + 8 = 60
S12 = 16 + 14 + 12 + 8 + 6 + 4 + 2 + 0 – 2 – 4 – 6 = 60
2
4.
If the 9th term of an A. P. is zero then prove that the 29th term is double
the 19th term.
(4 marks)
Given : t 9
= 0
Prove : t 29
= 2t19
Proof :
tn
= a + (n – 1) d
 t9
= a + (9 – 1) d
 0
= a + 8d
 a + 8d = 0
.......(i)
t 29
= a + (29 – 1) d

t 29
= a + 28d

t 29
= a + 8d + 20d

t 29
= 0 + 20d
[From (i)]

t 29
= 20d
.......(ii)
Similarly, t19 = a + (19 – 1)d

t 19
= a + 18d

t 19
= a + 8d + 10d

t 19
= 0 + 10d
[From (i)]

t 19
= 10d
......(iii)
Dividing (ii) by (iii),
t29
t19
t29
t19


5.
Sol.
=
=
t 29
=
Hence proved.
20d
10d
2
2t19
If ‘A’ and ‘G’ are the A. M. and G. M. between two numbers then prove
(5 marks)
that the numbers are A + (A + G) (A – G) .
Let the two number be x and y.
 A and G are A.M. and G.M. of x and y
x+y

A
=
.... (i)

G
=
G2
=
y
=
Now, 2A = x + y
S C H O O L S E C TI O N
2
xy
xy
G2
x
.... (ii)
45
MT
ALGEBRA






=

x
=

x
=

x
=


=
–b
b2 – 4ac
y
....(iii)
=
2A – x
x
2
G
=
2Ax – x2
2
2
x – 2Ax – G
=
0
Comparing with ax2 + bx + c = 0 we get, a = 1, b = –2A, c = G2
b2 – 4ac =
(–2A)2 – 4 (1) G2
2
b – 4ac =
4A2 – 4G2
=
4 (A2 – G2)
=
4 (A + G) (A – G)
x

2A – x
G2
EDUCARE LTD.
2a
– (– 2A) 
4 (A + G) (A – G)
2 ×1
2A  2 (A + G) (A – G)

2 A
2
(A  G) (A – G)

2
x =
A + (A + G) (A – G)
Substituting (iv) in (iii),
y
=
2A – [A  (A + G) (A – G) ]
y
=
2A – A  (A + G) (A – G)
y
=
A  (A + G) (A – G)
......(iv)
 The two numbers are A  (A + G) (A – G)
6.
Sol.
46
Find the sum of all terms given in a sequence 1, 2 + x, 3x2,4 + x3, ...., n + xn–1.
(5 marks)
For the given sequence 1, 2 + x, 3 + x2, 4 + x3 ......, n + xn–1
Sum of all terms of sequence
S = 1 + 2 + x + 3 + x2 + 4 + x3 + ...... + n + xn–1
 S = (1 + 2 + 3 + ..... + n) + (x + x2 + x3 + ...... + xn–1)
Consider 1 + 2 + 3 + ....... + n
These terms are in A.P. with a = 1, d = 1, tn = n
 Sum of all terms of A.P. with n terms
n
=
[t + tn]
2 1
n
=
[1 + n]
2
n  n2
......(i)
=
2
2
3
n–1
Consider, x + x + x + ..... + x
These terms form a G.P. with a = x, r = x
Sum of all terms of G.P. with n – 1 terms
a  r n – 1 – 1
=
r –1
x  x n – 1 – 1
=
x –1
S C H O O L S E C TI O N
MT
EDUCARE LTD.
ALGEBRA
 Sum of all terms of given sequence
= Sum of all terms of A.P. + sum of all terms of G.P.
n2  n x (x n – 1 – 1)

=
2
x –1
7.
If m times the mth term of an A. P. is equal to n times its nth term then
(4 marks)
show that the (m + n)th term of the A. P. is zero.
Given : mtm = nt n
Prove : tm+ n = 0
Proof :
t n = a + (n – 1) d
.... (i)
.... (ii)
Similarly tm = a + (m – 1) d
Similarly tm+ n= a + (m + n – 1) d
.... (iii)
 m [a + (m – 1) d]
= n [a + (n – 1) d]
 m [a + md – d]
= n [a + nd – d]
 am + m2d – md
= an + n2d – nd
 am – an + m2d – n2d – md + nd
= 0
 a (m – n) + d (m2 – n2) – d (m – n)
= 0
 a (m – n) + d (m + n) (m – n) – d (m – n) = 0
Dividing through at by m – n, we get
a + d (m + n) = d = 0
 a + d [m + n – 1] = 0
 a + (m + n – 1) d = 0
 tm+n + n = 0
[From (iii)]
Hence proved.
8.
Sol.
9.
Sol.
For a sequence Sn = n(3n + 2), find tn Examine whether the sequence is
(4 marks)
A. P or G. P.
S n = n (3n + 2)
 S 1 = 1 (3 (1) + 2) = 1(5)
= 5
= 16
S 2 = 2 (3 (2) + 2) = 2 (8)
= 33
S 3 = 3 (3 (3) + 2) = 3 (11)
= 56
S 4 = 4 (3 (4) + 2) = 4 (14)
= 5
t1 = S1
= 16 – 5
= 11
t 2 = S2 – S1
= 33 – 16 = 17
t 3 = S3 – S2
= 56 – 33 = 23
t 4 = S4 – S3
Now,
= 6
t2 – t1 = 11 – 5
t3 – t2 = 17 – 11 = 6
t4 – t3 = 23 – 17 = 6
 The difference between two consecutive terms is 6 which is a
constant, the sequence is an A.P.with a = 5, d = 6
We know for A.P. tn = a + (n – 1) d
 t n = 5 + (n – 1) 6
 t n = 5 + 6n – 6
 t n = 6n – 1
4 n – 3n
For a sequence Sn =
, find tn Examine whether the sequence is
3n
(5 marks)
A. P or G. P.
n
n
4 –3
Sn =
3n
1
4–3
1
4 – 31
S1 =
=
=
1
3
3
3
S C H O O L S E C TI O N
47
MT
ALGEBRA
S3 =
42 – 32
32
3
4 – 33
33
t1
=
S1
=
t2
=
S2 – S1
=
 t2
=
 t2
=
t3
=
 t3
=
 t3
=
S2 =




Now,
t2
t1 =
t2
t1 =
t2
t1 =
t3
t2 =
t3
t2 =
t3
t2 =
=
=
16 – 9
=
9
64 – 27
=
27
1
3
1
7
= –
3
9
EDUCARE LTD.
7
9
37
27
7
3
–
9
9
4
9
S3 – S2 =
37
7
–
27
9
37
21
–
27
27
16
27
4
1
÷
9
3
4
3
×
9
1
4
3
16
4
÷
27
9
16
9
×
27
4
4
3
 The ratio of two consecutive terms is
sequence is a G.P. with a =
4
which is a constant, the
3
1
3
4
3
We know for G.P.
 tn =
ar n–1
r
tn
=
=
1  4
 
3  3
n–1
  
48
S C H O O L S E C TI O N
S.S.C.
Marks : 30
CHAPTER 1 : Arithmetic Progression And Geometric Progression
ALGEBRA
Duration : 1 hr.
Q.1. Atttempt any TWO of the following :
(i)
For sequence, find the next four terms. 192, – 96, 48, – 24, ....
(ii)
State whether following list of numbers is an Arithmetic Progression ?
Justify. 1, 4, 7, 10, ....
(iii)
If 25 is the arithmetic mean between x and 46, then find x.
Q.2. Attempt any TWO of the following :
(i)
Find t11 from the following A.P. 4, 9, 14, ..... .
(ii)
Find the ninth term of the G.P. 3, 6, 12, 24, . ...
(iii)
Find the first three terms of the sequence for which S n is given
below : Sn = n2 (n + 1)
Q.3. Attempt any TWO of the following :
(i)
For an A. P. if t4 = 12, and d = – 10, then find its general term.
(ii)
Find the sum of the first n natural numbers and hence find the
sum of first 20 natural numbers.
(iii)
Find the ninth term of the G.P. 3, 6, 12, 24, . ...
Q.4. Attempt any TWO of the following :
(i)
A village has 4000 literate people in the year 2010 and this number
increases by 400 per year. How many literate people will be there
till the year 2020 ? Find a formula to know the number of literate
people after n years ?
2
4
6
8
... 2 ...
(ii)
If S3 = 31 and S6 = 3906 then find a and r.
(iii)
Find the indicated sum for the following Geometric Progression 2,
– 4, 8, –16, ... Find S9 and S12.
Q.5. Attempt any TWO of the following :
(i)
In a school, tree plantation on Independence day was arranged.
Every student from I standard will plant 2 trees, II standard students
will plant 4 trees each, III standard students will plant 8 trees
each etc. If there are 5 standard, how many trees are planted by
the student of that school ?
(ii)
Find three consecutive terms in a G.P. such that the sum of the
2nd and 3rd term is 60 and the product of all the three is 8000.
(iii)
Find three consecutive terms in a G.P. such that the sum of the
first two terms is 9 and the product of all the three is 216.
Best Of Luck 
10
S.S.C.
Marks : 30
CHAPTER 1 : Arithmetic Progression And Geometric Progression
ALGEBRA
Duration : 1 hr.
Q.1. Atttempt any TWO of the following :
(i)
For the sequence, find the next four terms. 3, 9, 27, 81, .....
(ii)
Write the first five terms of the following Arithmetic Progression where,
the common difference ‘d’ and the first term ‘a’ are given : a = 6, d = 6.
(iii)
If x + 3 is the geometric mean between x + 1 and x + 6 then find x.
Q.2. Attempt any TWO of the following :
(i)
Find the eighteenth term of the A. P. : 1, 7, 13, 19, .....
(ii)
Write down the first five terms of the geometric progression which
has first term 1 and common ratio 4.
(iii)
Find the first three terms of the sequence for which S n is given
n (n +1) (2n +1)
below :
6
Q.3. Attempt any TWO of the following :
(i)
Find tn for an Arithmetic Progression where t3 = 22, t17 = – 20.
(ii)
Find the sum of all odd natural numbers from 1 to 150.
(iii)
Write down the first five terms of the geometric progression which
has first term 1 and common ratio 4.
Q.4.
(i)
Attempt any TWO of the following :
Neela saves in a ‘Mahila Bachat gat ‘ Rs.2 on the first day, Rs.4 on
the second day, Rs.6 on the third day and so on. What will be her
saving in the month of February 2010 ?
2
4
6
8
(ii)
If S6 = 126 and S3 = 14 then find a and r.
(iii)
Find the indicated sum for the following Geometric Progression 1,
2 , 2, ... Find S10.
Q.5. Attempt any TWO of the following :
(i)
Find three consecutive terms in a G.P. such that the sum of the
first two terms is 9 and the product of all the three is 216.
(ii)
If the arithmetic mean and the geometric mean of two numbers are
in the ratio 5 : 4 and the sum of the two numbers is 30 then find
these numbers.
(iii)
In a school, tree plantation on Independence day was arranged.
Every student from I standard will plant 2 trees, II standard students
will plant 4 trees each, III standard students will plant 8 trees each
etc. If there are 5 standard, how many trees are planted by the
student of that school ?
Best Of Luck 
10