Question

If the 4th and the 9th terms of g.p is 54 and 13122 respectively, then its 2nd term is

Answer 1

Hello!

$\sf a_{n} = ar^{(n-1)}$

$\bf{a} : \sf First \:term$
$\bf{r} : \sf Common \:ratio$

Then,
$\sf a, ar, ar^{2}, ar^{3}, ar^{4}...$

$\star \: \sf Fourth \:Term:$ $\sf ar^{3} = \green{54}$

$\star \: \sf Ninth\: Term:$ $\sf ar^{8} = \green{13122}$

Divide the 9th term by the 4th term: 

$\dfrac{\sf ar^{8}}{\sf ar^{3}}$ = $\dfrac{13122}{54}$

$\implies \sf r^{5} = 243$

$\implies \sf r = 243^{\tfrac{1}{5}} \implies \sf \boxed{\bf{r} = \blue{3}}$

$\sf Now\: :$  

$\sf ar^{3} = 54$ 

$\implies \sf a(3^{3}) = 54$

$\implies \sf 27a = 54$

$\implies \sf a = \dfrac{54}{27} \implies \sf \boxed{\bf{a} = \blue{2}}$

$\bf{ANSWER}$ : 

$\sf a_{n} = 2 \times 3^{(n-1)}$

$\sf 2, \red{6}, 18, 54, 162, 486...$

Hence,
$\bf{The\: second\: term \:of\: G.P.\: is\: \red{\bf{6}}}$

Cheers!