untitled

Solution : The first term of the series a 2 , common ratio 2.
2

4

r

? The given series is a geometric series.
We know that the nth term of geometric series =arn^1
? 10 th term of the series = 2 u 201 ^1
= 2 u 29 0124
Example 7. What is the general term of the series 128  64  32 ?

Solution : The first term of the series a 128 ,common ratio.
2

1

128

64

r

? It is a geometric series.
We know that the general term of the series = arn^1

Hence, the general term of the series =.
2

1

2

1

2

2

1 17 8

(^17)
2
1
(^128)   

u ̧
¹
·
̈
©
§
n n n
n
Example 8. The first and the second terms of a geometric series are 27 and 7. Find
the 5th and the 1 0 th terms of the series.
Solution : The first term of the given series a 27 , the second term is 9.
Then the common ratio.
3
1
27
9
r
? The 5th term =
3
1
27 3
27 1
3
(^512714)
u
u ̧ u
¹
·
̈
©
ar  §
and the 1 0 th term =.
729
1
3
1
3 3
3
3
27 1
3 6 6
10 1 9 3
u
u ̧
¹
·
̈
©
ar  §
Determination of the sum of a Geometric series
Let the first term of th e geometric series be a, common ratio r and number of terms
n. If Sn is the sum of n terms,
Sn aarar^2 arn^2 arn^1 (i)
and r.Sn arar^2 ar^3 arn^1 arn [multiplying (i) by r] (ii)
Subtracting, SnrSn aarn
or, Sn( 1 r) a( 1 rn)
? ,
1
( 1 )
r
a r
S
n
n 

when r 1
Again, subtracting (ii) form (i) we get,
rSnSn arna or, Sn(r 1 ) a(rn 1 )
i.e., ,
( 1 )
( 1 )


r
ar
S
n
n when r!^1.
Observe : If common ratio is r 1 , each term a