ID : us-10-Arithmetic-Progressions [1]
Grade 10
Arithmetic Progressions
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Answer t he quest ions
(1)
T he sum of f irst 9 terms of an arithmetic progression is -234 and sum of its f irst 20 terms is 1400. What is the sum of the f irst 45 terms of this AP?
(2)
Find number of terms in the arithmetic progression 10, 17, 24, 31, ... such that sum of the
series is 4515?
(3)
T he nth term of an arithmetic progression is given by the equation -2 + 9n. What is the sum of
the f irst 30 terms of this AP?
(4) What is the value of the f ollowing series:
8 + 14 + 20 + 26 + ... + 98?
(5)
T he sum of f irst 5 terms of an arithmetic progression is 65. T he sum of the f irst 10 terms is
255. What is the sum of the f irst 15 terms of the AP?
(6) T he sum of the f irst three terms of an AP is 12, and their product is 504. What is the value of
the 13th term?
(7) T he sum of the 7 th element and the 15th element of an arithmetic progression is 198. T he sum
of the 10th and the 18th element is 252. What is the value of the 33rd term?
Choose correct answer(s) f rom given choice
(8)
If the sum of the f irst n terms of an arithmetic progression is given by the equation 24n + 4n 2,
then what is the 17 th element of this AP?
a. 156
b. 164
c. 148
d. 140
(9) Choose the arithmetic progression f rom the list below:
a. 6, -8, -22, -36
b. 7, -8, -21, -36
c. 6, -8, -21, -36
d. 6, -22, -36, -50
(10) If the 85th element of an arithmetic progression is 1197 and the 50th element is 707, then what
is the 5th element?
a. 77
b. 49
c. 63
d. 91
(11) What is the sum of all three-digit numbers that are divisible by 23?
a. 21527
b. 21528
c. 21531
d. 21530
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ID : us-10-Arithmetic-Progressions [2]
(12)
In an arithmetic progression, the value of the 13th element is
1
, and the value of 25th
25
element is
1
. What is the value of 325th element?
13
a. 2
b. 0
c. 1
d. 3
(13) A sequence of numbers that are in arithmetic progression starts with 17 and ends with 224.
What is the sum of the 11th term f rom the beginning of the sequence and the 11th term f rom
the end of the sequence?
a. 239
b. 243
c. 241
d. 240
(14) In an arithmetic progression, if the value of the ath term is b, and the value of the bth term is a,
then what is the value of the nth term?
a. a - b - n
b. n - b - a
c. a + b - n
d. a - b + n
(15) Which term in the arithmetic progression 7, 12, 17, 22 will be 135 more than the 20th term?
a. 46
b. 49
c. 47
d. 50
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ID : us-10-Arithmetic-Progressions [3]
Answers
(1)
-7650
Step 1
Sum of f irst n terms of arithmetic progression,
Sn = (n/2)[2a1 + (n - 1 ) d]
Step 2
It is given that sum of f irst 9 terms is -234,
S9 = -234
⇒ (9/2)[2a1 + (9 - 1 ) d] = -234
⇒ 9a1 + 9(9-1)(d/2) = -234
⇒ 9a1 + 36d = -234 ......................... (1)
Step 3
It is also given that sum of f irst 20 terms is -1400,
S20 = -1400
⇒ (20/2)[2a1 + (20 - 1 ) d] = -1400
⇒ 20a1 + 20(20-1)(d/2) = -1400
⇒ 20a1 + 190d = -1400 ......................... (2)
Step 4
On solving Eq. (1) and (2), we get,
a1 = 6 and d = -8
Step 5
Now, sum of f irst 45 terms,
S45 = (45/2)[2a1 + (45 - 1 ) d]
⇒ S45 = (45/2)[2(6) + (45 - 1 ) (-8)]
⇒ S45 = (45/2)[-340]
⇒ S45 = -7650
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ID : us-10-Arithmetic-Progressions [4]
(2)
35
Step 1
T he given AP is, 10, 17, 24, 31, ...
Now, f irst term a = 10,
Dif f erence between successive terms d = 17 - 10 = 7
Step 2
We know that, Sn =
n
[2a + (n - 1)d]
2
Let's assume, the given AP have 'n' number of terms.
According to the question, Sn = 4515
T heref ore, 4515 =
n
[(2 × 10) + (n - 1)(7)]
2
⇒ 4515 =
n
[7n + (13)]
2
⇒ 9030 = [7n2 + (13n)]
⇒ 7n2 + 13n - 9030 = 0
Step 3
T he number of terms neither negative nor in f raction, the value of 'n' must be positive.
By solving above quadratic equation, we get:
n = 35
Step 4
T heref ore, this AP should have 35 terms.
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ID : us-10-Arithmetic-Progressions [5]
(3)
4125
Step 1
According to the question, an = -2 + 9n
T heref ore, a1 = 7,
a2 = 16,
a3 = 25
Now, the required AP: 7, 16, 25 .........
d = a2 - a1
= 16 - 7
=9
Step 2
T he sum of the f irst 30 terms of this AP(S30 ) =
30
30
[2a + (30 - 1)d]
2
[(2 × 7) + (30 - 1)(9)]
2
=
30
[(2 × 7) + (29) × (9)]
2
= 4125
Step 3
T hus, the sum of the f irst 30 terms of this AP is 4125.
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ID : us-10-Arithmetic-Progressions [6]
(4) 848
Step 1
From f irst f our terms, we can observe that this is an arithmetic progression with dif f erence
of 6
Step 2
If there are n terms in the arithmetic progression, its last term is given by,
T n = T 1 + (n-1)d
⇒ 98 = T 1 + (n-1)d
⇒ 98 = 8 + (n-1)6
⇒ n-1 = (98 - 8)/6
⇒ n-1 = 15
⇒ n = 16
Step 3
Now sum of arithmetic progression can be f ound using standard f ormula,
Sn = (n/2)[2T 1 + (n-1)d ]
⇒ S16 = (16/2)[2(8) + (16-1)(6) ]
⇒ S16 = (16/2)[ 106 ]
⇒ S16 = 848
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ID : us-10-Arithmetic-Progressions [7]
(5)
570
Step 1
Sum of f irst n terms of arithmetic progression,
Sn = (n/2)[2a1 + (n - 1 ) d]
Step 2
It is given that sum of f irst 5 terms is 65,
S5 = 65
⇒ (5/2)[2a1 + (5 - 1 ) d] = 65
⇒ 5a1 + 5(5-1)(d/2) = 65
⇒ 5a1 + 10d = 65 ......................... (1)
Step 3
It is also given that sum of f irst 10 terms is 255,
S10 = 255
⇒ (10/2)[2a1 + (10 - 1 ) d] = 255
⇒ 10a1 + 10(10-1)(d/2) = 255
⇒ 10a1 + 45d = 255 ......................... (2)
Step 4
On solving Eq. (1) and (2), we get,
a1 = 3 and d = 5
Step 5
Now, sum of f irst 15 terms,
S15 = (15/2)[2a1 + (15 - 1 ) d]
⇒ S15 = (15/2)[2(3) + (15 - 1 ) (5)]
⇒ S15 = (15/2)[76]
⇒ S15 = 570
(6) -51
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ID : us-10-Arithmetic-Progressions [8]
(7) 297
Step 1
If d is the dif f erence between consecutive terms, the nth term of arithmetic progression is,
T n = T 1 + (n-1) d
Step 2
It is given that sum of the 7 th element and the 15th element is 198
T 7 + T 15 = 198
⇒ T 1 + (7-1) d + T 1 + (15-1) d = 198
⇒ 2T 1 + 20d = 198 ............................(1)
Step 3
It is also given that sum of the 10th element and the 18th element is 252
T 10 + T 18 = 252
⇒ T 1 + (10-1) d + T 1 + (18-1) d = 252
⇒ 2T 1 + 26d = 252 ............................(2)
Step 4
On subtracting Eq. (1) f rom (2)
[2T 1 + 26d] - [2T 1 + 20d] = 252 - 198
⇒ (26 - 20) d = 54
⇒ (6)d = 54
⇒d=9
Step 5
On subtracting d in Eq. (1)
⇒ 2T 1 + 20(9) = 198
⇒ 2T 1 = (198) - (180)
⇒ T 1 = [(198) - (180)]/2 = 9
Step 6
T heref ore 33rd term,
T 33 = T 1 + (33-1) d
⇒ T 33 = 9 + (33-1)(9)
⇒ T 33 = 9 + (288) = 297
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ID : us-10-Arithmetic-Progressions [9]
(8)
a. 156
Step 1
For an arithmetic progression with f irst term = a1, and common dif f erence of successive
members = d, sum of f irst n terms is given by,
Sn = (n/2)[2a1 + (n-1)d ]
Step 2
It is given that sum of f irst n terms is 24n + 4n2,
Sn = 24n + 4n2
(n/2)[2a1 + (n-1)d ] = 24n + 4n2
⇒ (n/2)[2a1 + nd - d ] = 24n + 4n2
⇒ na1 + (d/2)n2 - (d/2)n = 24n + 4n2
⇒ (a1 - d/2)n + (d/2)n2 = 24n + 4n2 ...................... (1)
Step 3
On comparing f actors of n2 in Eq. (1)
d/2 = 4
⇒d=8
Step 4
On comparing f actors of n in Eq. (1)
a1 - d/2 = 24
⇒ a1 - d/2 = 24
⇒ a1 - 4 = 24
⇒ a1 = 24 + 4
⇒ a1 = 28
Step 5
Now we can f ind 17 th term using standard f ormula f or arithmetic progression,
an = a1 + (n-1) d
⇒ a17 = 28 + (17-1) 8
⇒ a17 = 28 + 128
⇒ a17 = 156
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ID : us-10-Arithmetic-Progressions [10]
(9) a. 6, -8, -22, -36
Step 1
In an arithmetic progression, dif f erence between consecutive terms is f ixed. Lets check this
f or all the given progressions
Step 2
Dif f erences in series 6, -8, -21, -36 are
(-8) - (6) = -14
(-21) - (-8) = -13
(-36) - (-21) = -15
Step 3
Dif f erences in series 6, -22, -36, -50 are
(-22) - (6) = -28
(-36) - (-22) = -14
(-50) - (-36) = -14
Step 4
Dif f erences in series 6, -8, -22, -36 are
(-8) - (6) = -14
(-22) - (-8) = -14
(-36) - (-22) = -14
Step 5
Dif f erences in series 7, -8, -21, -36 are
(-8) - (7) = -15
(-21) - (-8) = -13
(-36) - (-21) = -15
Step 6
It can be seen that dif f erence is f ixed only f or series (6, -8, -22, -36), theref ore (6, -8, -22,
-36) is an arithmetic progression
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ID : us-10-Arithmetic-Progressions [11]
(10) a. 77
Step 1
If d is the dif f erence between consecutive terms, the nth term of arithmetic progression is,
T n = T 1 + (n-1) d
Step 2
It is given that 85th element is 1197,
T 1 + (85-1) d = 1197
⇒ T 1 + 84d = 1197 ............................ (1)
Step 3
Similarly it is given that 50th element is 707,
T 1 + (50-1) d = 707
⇒ T 1 + 49d = 707 ............................ (2)
Step 4
On subtracting equation (1) f rom (2),
84d - 49d = (1197) - (707)
35d = 490
d = (490)/(35)
d = 14
Step 5
Replace d by 14 in equation (2)
⇒ T 1 + 49(14) = 707
⇒ T 1 = 707 - 49(14)
⇒ T 1 = 21
Step 6
T heref ore 5th term will be, T 5 = T 1 + (5-1) d
T 5 = 21 + (5-1) 14
T 5 = 77
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ID : us-10-Arithmetic-Progressions [12]
(11) b. 21528
Step 1
All numbers which are divisible by 23, f orms an arithmetic progression with dif f erences
between consecutive terms to be 23
Step 2
We know smallest three digit number is 100. On dividing 100 by 23 we get remainder of 8
T heref ore if we if we add remaining (23 - 8 = 15) to 100, resultant number (100 + 15 = 115)
will be f ully divisible by 23
T heref ore f irst number of the arithmetic progression is 115
Step 3
Similarly, largest three digit number is 999. On dividing 999 by 23 we get remainder of 10
T heref ore if we if we subtract 10 f rom 999, resultant number (999 - 10 = 989) will be f ully
divisible by 23
T heref ore last number of the arithmetic progression is 989
Step 4
If there are total N terms in series, Nth term is given by
T N = T 1 + (N-1)d
⇒ 989 = 115 + (N-1)(23)
⇒ 23(N - 1) = 989 - 115
⇒ 23(N - 1) = 874
⇒ N - 1 = 874/23
⇒ N - 1 = 38
⇒ N = 38 + 1
⇒ N = 39
Step 5
Now sum of arithmetic progression can be f ound using standard f ormula,
SN = (N/2)[T 1 + (N-1)d]
⇒ SN = (39/2)[2×115 + (39-1)(23)]
⇒ SN = (39/2)[230 + 874]
⇒ SN = (39/2)[1104]
⇒ SN = 21528
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ID : us-10-Arithmetic-Progressions [13]
(12) c. 1
Step 1
We know that, an = a + (n - 1)d,
Where, n = number of elements in an AP(arithmetic progression),
a = f irst element of an AP,
d = common dif f erence.
Step 2
According to the question, a13 =
1
,
25
a13 = a + (13 - 1)d
1
⇒ a + (13 - 1)d =
25
1
⇒ a + 12d =
25
1
⇒a=
- 12d -----(1)
25
Step 3
1
a25 =
,
13
a25 = a + (25 - 1)d
By putting the value of 'd' f rom equation (1), we get:
1
1
- 12d + 24d =
25
13
⇒ 12d =
1
-
13
12
⇒d=
1
25
-----(2)
3900
Step 4
By putting the value of 'd' in equation (1), we get:
a=
1
25
12
- 12 ×
3900
12
=
3900
Step 5
Now, a325 = a + (325 - 1)d
12
=
+ 324 ×
3900
=1
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12
3900
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=1
ID : us-10-Arithmetic-Progressions [14]
Step 6
T hus, the value of 325th element is 1.
(13) c. 241
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ID : us-10-Arithmetic-Progressions [15]
(15) c. 47
Step 1
If you inspect this series, you will f ind that dif f erence between consecutive terms is 5
Step 2
Since dif f erence between consecutive terms (i.e. distance of 1) is 5, dif f erence between
terms with distance n will be 5n. i.e.
T a - T b = 5(a-b)
Step 3
Lets required term be mth term,
T m - T 20 = 135
⇒ (m - 20)5 = 135
⇒ m - 20 = 135/5
⇒ m - 20 = 27
⇒ m = 27 + 20
⇒ m = 47
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