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 .

Hope helped!

Answer 2

Step-by-step explanation:

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× 2^2 = 20

a = 5

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

Sum of first 9 terms = 5(1-2^9)/1-2

= 5 (1-512 ) /-1

= -5(-511)

=2555

= 5110.

corrected value