Section Check In – 1.04 Sequences and Series
Questions
1.
Find the first 4 terms, in ascending powers of x , of the binomial expansion of  2  x  , giving
6
each term in its simplest form.
2.
Find the coefficient of x 3 in the expansion of  3  2x  .
3.*
A sequence of numbers u1 , u2 , u3 , is given by u1 1 , un 1  2  un   3 .
5
2
Find u 2 , u3 , and u4 .
4.*
Find the binomial expansion of  3  2x 
4
in ascending powers of x , as far as the term
in x 2 . Give each coefficient in its simplest form and state the values of x for which the
expansion is valid.
5.
Find the coefficient of x 2 in the expansion of  x  2  2 x  3 .
4
12
6.*
Find
3 2
r
.
r 1
7.
(a) Write down the first 3 terms, in ascending powers of x , of the binomial expansion of
1 ax 
7
, where a is a non-zero constant.
(b) Given that in the expansion of 1  ax  , the coefficient of x 2 is nine times the coefficient
7
of x , find the value of a .
8.*
The second and fifth terms of a geometric sequence are 32 and 0.5 respectively.
For this series, find
(a) the common ratio and the first term of the sequence,
(b) the sum of the first five terms,
(c) the sum to infinity giving your answer to three decimal places.
9.*
An athlete is training for the London marathon. He runs 7 miles on day 1 and then increases
his run by 0.8 miles each day until he can run a complete marathon of 26.2 miles.
(a) On which day of his training does he run the full 26.2 miles?
(b) How many miles does he run over the whole training period?
10.* A hat shop made a profit of £20 000 in 2015. A model for future trading predicts that profit will
increase each year in a geometric sequence so that in 2016 the profit will be £ 20 000 r .
(a) Give an expression for the expected profit in the year 2020.
Given that r 1.07
(b) find the year that the profit will first exceed £ 30 000,
(c) find the total profit made by the hat shop in the years 2015 to 2020. Give your answer to
the nearest hundred pounds.
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Extension
If the first two terms of a sequence form an arithmetic sequence and also form a geometric
sequence, what is the relationship between a , d and r ?
Is it possible for the first three terms of a sequence to form an arithmetic sequence and also form a
geometric sequence? If so, what can you say about d and r ?
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Worked solutions
1.
 6
 6
 6
26    25 x    24 x 2    23 x3  64  192 x  240 x 2 160 x3
1
 2
 3
2.
5 2
3
3
  3  2 x  Coefficient of x is 1098 720
31
 
3.
u1  1 , u2  2 12  3  5 , u3  2  52  3  53 , u4  2  532  3  5621
4.
5.
2
  2   4 1 
 2   4  5  2  
x 
 3  2 x   3 1  x   1   4   x  

81 
2
 3 
 3  
  3 
3
1 8
40  1
8
40 2
 1  x  x 2   
x
x . This is valid for x 
2
81  3
9
729
 81 243
4
 2 x  3
4
3
2
2
3
4
  2 x   4   2 x    3  6   2 x    3  4   2 x    3    3 
4
 16 x 4  96 x3  216 x 2  324 x  81
 x  2 2x  3   x  2 81  324 x  216 x2 
4
 81x  162 324 x 2  648 x  432 x 2 
Coefficient of x 2 is 432 – 324 1
08
6.
u1  6 , u2  12 , u3  24
a  6 , r  2 , n  12
6 1  212 
S12 
 24570
1 2
7.
(a) 1  7ax  21a 2 x 2
(b) 21a 2  9 7 a , 21a  63 , a  3
8.
u5  0.5
1
1
1
4
3
3
(a) 32  ar ,  ar  32r , r 
, r  . a  128
2
64
4
5
 1 
128 1    
 4 

  170.5
(b) S5 
1
1
4
a
128

 170.667 to 3 d.p.
(c) S 
1 r 1 1
4
u2  32
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9.
a  7 , d  0.8
(a) un  26.2  7  0.8  n  1
19.2  0.8 n 1
24   n  1
n  25 days
25
(b) S25   2 7  24 0.8   415 miles
2
10.
(a) 20000  r 5
(b) 20000 1.07 n  30000
1.07 n  1.5
n log1.07  log1.5
n 5.99
6 years later will be the year 2021
(c) a  20000 , r  1.07 , n  6
S6 

20000 1  1.07 
1  1.07
6
  143066
£143 100 to the nearest hundred
Extension
If the first two terms form an arithmetic and a geometric sequence
ar  a  d
 d
ar  a 1  
 a
d
Therefore r  1 
a
If the first three terms form an arithmetic and a geometric sequence
ar  a  d and ar 2  a  2d
ar  a  d so r 
ad
a
ad 
ar 2  ar r  r  a  d   
  a  d 
 a 
But ar  a  2d so
2
a  d 
a  d 
a
2
 a  2d
2
a 2  2ad  d 2
 a  2d
a
a
a 2  2ad  d 2  a 2  2ad
Therefore d 2  0 so d  0
ad a0
r

1
a
a
So

In this case, all terms of the sequence must be equal
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