Question

In a GP if T3 = 18 and T6 = 486, find T10

Answer 1

The 10th term is 39366

Explanation:

The n-th term in a geometric progression is

`T_n = a {r}^(n - 1)`

The third term is

`T_3 = a {r}^(2) = 18`

The 6th term is

`T_6 = a {r}^(5) = 486`

Let us take the ratio:

`(T_6)/(T_3) = \frac{a {r}^(5) }{a {r}^(2) } = (486)/(18) = 27`

This means that:

`{r}^(3) = 27`

`r = \sqrt[3]{27} = 3`

Put this into the 3rd term

`a * {3}^(2) = 18`

`9a = 18`

`a = 2`

The 10th term is:

`T_ {10} = a {r}^9`

`T_ {10} = 2 * {3}^(9)= 39366`