Question

which term of GP 3,3√3,9.....is 243?with step by step answer

Answer 1

nth term of a GP =
$a \times {r}^{n - 1 }$
$r = \sqrt{3}$
$3 \times { \sqrt{3} }^{n - 1} = 243 \\ { \sqrt{3} }^{n - 1} = 8 1$
${ \sqrt{3} }^{n - 1} = { \sqrt{3} }^{8}$
n-1=8
n=8+1=9

Answer 2

Answer:

9th term of GP is 243.        

Step-by-step explanation:

Given : GP 3,3√3,9.....

To find : Which term of GP is 243 ?

Solution :

GP 3,3√3,9.....

The first term of GP is 3.

The common ratio is $r=\frac{3\sqrt3}{3}=\sqrt{3}$

The nth term of the GP is given by,

$T_n=a \times  {r}^{n - 1 }$

Here, $T_n=243$

$243=3\times  {\sqrt{3}}^{n - 1 }$

$81={\sqrt{3}}^{n - 1 }$

${\sqrt{3}}^{8} ={\sqrt{3}}^{n - 1 }$

Comparing the base power is equal,

$8=n-1$

$n=8+1$

$n=9$

Therefore, 9th term of GP is 243.