Question

Find the sum of the first six terms of a geometric progression .​

Answer 1

Question:

Find the sum of the first six terms of a geometric progression.

1,3,9,....

Answer:

`S_6 = 364`

Explanation:

For a geometric progression, the sum of n terms is:

`S_n = (a(r^n - 1))/(r - 1)`

In the given sequence:

`a = 1`

`r = 3/1 =3`

`n = 6`

So:

`S_n = (a(r^n - 1))/(r - 1)`

`S_6 = (1 * (3^6 - 1))/(3 - 1)`

`S_6 = (3^6 - 1)/(2)`

`S_6 = (728)/(2)`

`S_6 = 364`