Question

6. The second term of a geometric series is 3 and the common ratio is . 4\5 Find the sum of first 23 consecutive terms in the given geometric series.

Answer 1

Let first term is a
Given, second term = 3
Common ratio , r = 4/5
so, second term = ar²⁻¹ = ar
3 = a × 4/5 ⇒a = 15/4

Now, sum of first n terms , Sn = a[1 - rⁿ]/(1 - r) [ when r < 1 ]
S₂₃ = (15/4) [1 - (4/5)²³]/(1 - 4/5)
= (15/4) [ 1 - (4/5)²³]/(1/5)
= (75/4)[ 1 - (4/5)²³ ]

Answer 2

HELLO DEAR,

Let first term be a

GIVEN THAT:-

second term = 3 , Common ratio , r = 4/5

second term = ar²⁻¹ = ar

3 = a × 4/5

a = 15/4

Now,

sum of first n terms , $\mathbf{S_n = a[1 - r^n]/[1 - r]}$ (if r < 1 )


S₂₃ = (15/4) [1 - (4/5)²³]/(1 - 4/5)

= (15/4) [ 1 - (4/5)²³]/(1/5)

= (75/4)[ 1 - (4/5)²³ ]


I HOPE ITS HELP YOU DEAR,
THANKS