Question

find the sum of the first 9 terms of a geometric series whose third term and 7 terms are 20 and 320 respectively​

Answer 1

We know that nth term of a geometric progression is a*r^n-1.

Now third term = ar² = 20

Seventh term = ar^6= 320

Dividing the above, we get

r^4 =16

r = 2.

Now, ar² = 20

a = 10.

We know that sum of n terms in geometric progression is a (1 - r^n)/1-r.

Sum of first 9 terms = 10(1-2^9)/1-2

= 10 (1-512) / -1

= -10(-511)

= 5110.