Example 1:
Find the 10th term and the nth term for the
sequence 7, 10, 13, … .
Solution:
a7
d 3
U10 = 7  10  1 3 
 34
Un= 7   n  1 3 
 3n  4
1
Example 2
Find the three numbers in an arithmetic
progression whose sum is 24 and whose
product is 480.
Solution
Let the three numbers be (a-d), a, (a+d)
where d is the common difference.
(a-d)(a)(a+d)=480
(a-d) +a + (a+d) = 24
Substituting a =8,
3a = 24
we get
a=8
d  2
2
When d = 2, required numbers are 6, 8, 10
When d = -2 , required numbers are 10, 8, 6
3
Example 3
The sum of the first eighteen terms of an
arithmetic series is -45 and the eighteenth
term is also -45.
U 18   45
Find the common difference and the sum of
the first hundred terms.
Solution
Sn 
n
2
2 a   n  1 d 
S 18   45 
18
[ 2 a  (18  1) d ]
2
 18 a  153 d   45
--------- (1)
4
U 18   45
a  18  1 d   45
a  17 d   45 ------------ (2)
Solving equation 1 and 2, we get
a  40
d  5
5
Example 4
Find the sum of the positive integers which
are less than 500 and are not multiples of
11.
Solution
Sum of integers
Required sum= from 1 to 499
(SI)
S I  1  2  3  4  .......... ..  499
499
1  499 

2
 124750
Sum of
multiples
of 11(SII)
6
S II  11  22  33  44  .......  495
 11 1  2  3  4  ........  45 
 45

1  45 
 11
 2

Sn 
n
a  l 
2
 11385
Required sum = SI-SII
=124750-11385
=113365
7
Example 5
S 10  145
The sum of the first 10 terms of an A.P. is
145 and the sum of the next 6 terms is 231
Find (i) the 31st term , and U 31  a  30 d
(ii) the least number of terms required
for the sum to exceed 2000.
S10  U  U
11
12  U 13  U 14  U 15  U 16  231  145
S16  16 2 a  15 d 
2
8
Solution
S16 
16
2
2 a  15 d   145
 231
82 a  15 d   376
2 a  15 d  47
(1)
S 10  145
10
2
2 a  9 d   145
2 a  9 d  29
(2)
9
(1)(2):
6 d  18
d 3
From (2): a 
29  9  3 
1
2
U 31  a  30 d  1  30  3   91
(ii)Find the least number of terms required
for the sum to exceed 2000.
Let n be the least number of terms
required for S n  2000
10
n
2
n
2
2 a   n  1 d  
2000
2 1    n  1 3  
2000
n 3 n  1  4000
3n
2
 n  4000  0
2
Consider 3 n  n  4000  0
1  1  4  3   4000 
n
 36 . 7 or  36 . 3
6
11
For
3n
2
 n  4000  0

+
n   36 . 3 or n  36 . 7
 36 . 3
+
36 . 7
Since n must be a
positive integer, n > 36.7
Hence, the least number of terms required
is 37.
12
Example 6
Given that the fifth term of a geometric
progression is
81
4
and the third term is 9.
Find the first term and the common ratio if all
the terms in the G.P. are positive.
ar
4

81
(1)
4
ar
2
9
(2)
13
(1)(2):
ar
ar
4
2
r
2

81

4

1
9
9
4
Since all the terms in the G.P. are positive,
r>0
3
r 
2
2
From (2): a  3   9
2
a4
14
Example 7
Three consecutive terms of a geometric
x 1
x
progression are 3 , 3
and 81. Find the
value of x. If 81 is the fifth term of the
geometric progression, find the seventh
term.
x 1
r 
3
3
3

x
3
x 1
3
x
81

3
3
x 1
4
x 1
15
3 x
33
1 3 x
81
r 
x2
3
x 1

3
4
3
3
3
Given 81 is the fifth term, find the seventh
term:
U5= ar4 = 81
a 3   81
4
a
81
3
U7 =
ar6
1
4
 1( 3 )  729
6
16
Example 8
Find the sum of the first eight terms of the
series 3  2  4  8  .........
3
Soln:
2

a 1 r
1 r
8

U2
8

2
3 1    
 3  


2
1
3
2
U3
2 U4 2

 ,
 ,

a  3, r 
3
U1 3 U 2 3 U 3 3
G.P. with
S8 
9
17
8

2
3 1    
 3  


2
1
3
256 

 3 1 
3
6561 


6305
729
18
Example 9
a 1
A geometric series has first term 1 and
S5
the common ratio r, where r 1, is
positive. The sum of the first five terms is
twice the sum of the terms from the 6th to
15th inclusive. Prove that
r
5

1
( 3  1).
U 6  U 7  .....  U 15
2
 U 1  U 2  U 3  U 4  U 5   U 6  U 7  ........  U 15
 U 1  U 2  U 3  U 4  U 5 
 S15  S 5
19
Solution
S 5  2  S15  S 5 
Given
3 S 5  2 S 15
 a ( r 5  1) 
 a ( r 15  1) 
 3
  2

 r  1 
 r  1 
Since r  1 and a = 1, we have
5
3 ( r  1)  2 ( r
2r
15
 3r
5 3
5
15
 1)
1 0
5
2(r )  3r  1  0
20
Let x  r 5 .
So the equation becomes
3
2x  3x  1  0
Let f(x) = 2 x 3  3 x  1
Since f(1) = 2 – 3 + 1 = 0
(x –1) is a factor of f(x).
3
2 x  3 x  1  ( x  1)( 2 x
2
 kx  1)
Comparing coefficient of x:  3   1  k
 k 2
3
2 x  3 x  1  ( x  1)( 2 x
2
 2 x  1)
21
( x  1)( 2 x
Hence,
2
 2 x  1)  0
 x  1 or 2 x
x
2
 2x 1  0
2
4  4 ( 2 )(  1)
2(2)
5
 r  1 or r
 r  1
5

2
12

22 3
4

1
4
( 3  1) or
2
5
1
(  3  1)
2
(r<0)
1
Since r  1 and r > 0,  r  ( 3  1)
2
22
Sum to infinity, S (or S  ) =
a
1 r
The sum to infinity exists (the series converges
or series is convergent) provided r  1
Example 10
Determine whether the series given below
converge. If they do, give their sum to
infinity.
G.P. with r  2 >1
(a) 2  4  8  16  ........
Does not converge
23
(b) 1 
1

4


16
S 

1
1
64
a
1 r
 .......... .
G.P. with
r 
1
r 
<1
1
4
4
Series converges
1
 1
1  
 4
4
5
24
(c) 3 
9
2

27
4
 .......... ..........G.P.
3
with r 
2
3
>1
r 
2
Does not converge
25
Example 11
A geometric series has first term a and the
common ratio 1 .
2
Show that the sum to infinity of the geometric
progression is a ( 2  2 ) .
Solution S  

a
1 r

a (2 
2 1
a
1 

1 

2


2 )  a (2 
a 2
2 1
2)

2 1
2 1
(shown)
26
Example 12
A geometric series has first term a and
common ratio r. S is the sum to infinity of the
series, T is the sum to infinity of the evennumbered terms (i.e.U 2  U 4  U 6  ....) of the
series. Given that S is four times the value
2
common
ratio
r
of T, find the value of r.
S 
a
1 r
T  ar  ar 3  ar 5  .........

ar
1 r2
Given S = 4T, we have
a
1 r
4
ar
1 r
2
27

1
1 r

 1
4r
(1  r )( 1  r )
4r
1 r
 1  r  4r
r 
1
3
28