Example 13
The first two terms of a geometric
progression are 3 and –2. Find the least
value of n for which the difference
between sum of the first n terms and sum
to infinity is within 2% of the sum to
infinity.
2
2
S

S

S
a  3 and r  
n
100
3
a
a(1  r n )
2
a


(
)
1 r
1 r
100 1  r
1




n
3
3(1  r )
2 
3



2
2
2
100 



1 
1 



 1 
3
3
3









9 9
2  9
n

1 r

 
5 5
100  5 
9 n
2  9
r 
 
5
100  5 
 2
  
 3
n
2

100
Note :
n
n
 2   2
 
3
 3
n
2
 
  
 3 2

 2


 3
n
 0.02
 2
 n l g
  l g 0.0 2
 3
Since
n  9.65
Hence the least value of n is 10. (ans)
3
Example 14
The sum of the first n terms of a series is
2
2n  n.
Find the fifth term. Prove that the series is
arithmetic and state its common difference.
Solution
2
Given S n  2n  n .
Therefore, U 5  S5  S4
 [2(5) 2  5]  [2(4) 2  4]
 19
4
To prove that the series is an A.P., prove
U n  U n1  cons tan t
2
S

2
n
n ,
Given n
S n1  2n  12  n  1
U n  Sn  Sn1
 (2n 2  n)  [2(n  1) 2  (n  1)]
2
2
 2n  n  (2n  4n  2  n  1)
 4n  1
U n1  4n  1  1
5
U n  U n1  (4n  1)  [4(n  1)  1]
 4n  1  4n  5  4 , (a cons tan t )
Thus, the series is arithmetic with common
difference 4.
6
Example 15
The sum of the first n terms of a series is
n 1
given by 6  2
. By finding an expression
n 1
3
Sn
for the nth term of the series, show that this
is a geometric series, and state the value of
the first term and the common ratio.
Solution
2n1
2( n1)1
U n  Sn  Sn1  [6  n1 ]  [6  (n1)1 ]
3
3
7

n 1
2
n 1
3
2

 
n
n2
3
n
n
22
2


 3n   3n 
   
 3   9 
   
 2n   2n 
 6   9 
 3n   3n 
   
n
2
 
U n  3 
 3
To show G.P., show
Un
 cons tan t
U n1
n
2

3 
Un
3


n 1
U n1
2

3 
 3
2
 (a cons tan t )
3
8
Therefore, the series is geometric with first term
 2
U1  3 
 3
2
2
and common ratio 
3
9
Example 16
The sum of the first 100 terms of an
Arithmetic Progression is 10, 000; the first,
second and fifth terms of this progression
are three consecutive terms of a Geometric
Progression. Find the first term, a , and the
non-zero common difference, d, of the A.P.
Solution
100
2a  100  1d   10000
S100 
2
2a  99d  200 ----(1)
a, a+d, a+ 4d are three consecutive terms
10
of a Geometric Progression
a  d a  4d
common ratio 

a
ad
a  d   a a  4 d 
2
a 2  2ad  d 2  a 2  4ad
d 2  2ad
d  2a ----(2)
Substituting (2) into (1), 2a  99(2a)  200
a 1
From (2),
d 2
11
Example 17
r 1
The rth term of a series is 3
 2r . Find the
sum of the first n terms.
Solution
Given
ur  3r 1  2r
The sum of the first n terms is:
S n  u1  u 2  u3  ...u n

 
 



Sn  311  2(1)  321  2(2)  331  2(3)  ... 3n1  2(n)
12

11
21
31
n1
Sn  3  3  3  ...  3

 2(1)  2(2)  2(3)  ... 2(n)

S n  1  3  9  ...3n1  21  2  3  ...  n
GP: a = 1, r = 3, n
terms


AP: a = 1, d = 1,
n terms
1 3 1
n


 2 1  n 
3 1
2

n
1

(3 n  1)  n (1  n)
2


1
3 n  1  2n  2n 2
2

13
Example 18
Each time that a ball falls vertically on to a
horizontal floor it rebounds to three-quarters
of the height from which it fell. It is initially
dropped from a point 4 m above the floor.
Find, and simplify, an expression for the total
distance the ball travels until it is about to
touch the floor for the (n+1)th time. Hence
find the number of times the ball has bounced
when it has traveled 24 m and also the total
distance it travels before coming to rest. (The
dimensions of the ball are to be ignored.)
14
2
3
 
  4 
 4
n
 3
 4
 4
4m
1st
2nd
 3
  4 
 4
3rd
nth (n+1)th
Total distance (in metres) that the ball travels
n
2
3
 3
 3

 2  4 
 4  2 4  2  4  
 4
 4
 4
n
 3   3  2  3 3
 3
 4  8         .......    
15
4
4
4
4










 3   3  n  
 1     
 4   4   
 4  8

3


 1  

 4




Re call

a 1 r
Sn 
1 r
n

  3 n 
 4  241    
  4 


n
3
 
 28  24 
 4
16
Hence find the number of times the ball has
bounced when it has traveled 24 m and
also the total distance it travels before
coming to rest Let the number of times the
ball has bounced be n.
Solution
n
3
 
4  24 
When the ball has
 4
traveled 24 m,
n
3
1
n



 3


28  24   24
4
6

 4
1
 3
n lg   lg
 3
Note : lg   0
6
 4
17
 4
1

lg 
6  Since n is an integer, least n=7.

n
3  Therefore, the ball has bounced

lg 
 4  7 times when it has traveled 24
m.
n  6.23
n
For the ball to come to
 3
As n  ,    0
rest, n  
 4
n
3
 
therefore, 28  24   28
 4
The ball travels 28 m before coming to
18
rest.