Question

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

Answer 1

Answer:

The 10th term is 39366

Step-by-step 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:

$\frac{T_6}{T_3} = \frac{a {r}^{5} }{a {r}^{2} } = \frac{486}{18} = 27$

This means that:

${r}^{3} = 27$

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

Put this into the 3rd term

$a \times {3}^{2} = 18$

$9a = 18$

$a = 2$

The 10th term is:

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

$T_ {10} = 2 \times {3}^{9}= 39366$