Question

The fourth term, the seventh term and the last term of a geometric progression are 10, 80 and 2560 respectively. Find its first term, common ratio and number of terms.

Answer 1

Let the first term of the G.P. be a and its common ratio be r.

4^th term  = t₄ = 10 `=>` ar³ = 10

7^th term = t₇ = 80 `=>` ar⁶ = 80

`(ar^6)/(ar^3) = 80/10`

`=>` r³ = 8

`=>` r = 2

ar³ = 10

`=>` a × (2)³ = 10

`=> a = 10/8 = 5/4`

Last term = I = 2560

Let there be n term in given G.P.

`=>` t_n = 2560

`=>` ar^{n – 1} = 2560

`=> 5/4 xx (2)^(n - 1) = 2560`

`=>` (2)^{n – 1} = 2048

`=>` (2)^{n – 1} = (2)¹¹

`=>` n – 1 = 11

`=>` n = 12

Thus, we have

First term = `5/4` common ratio = 2 and number of terms = 12