Question

how to solve the number series of 9,36,81,144,?

Answer 1

the answer would be 225

Answer 2

$Given\: series : 9, 36, 81, 144 , \red {?}$

/* Observe the pattern in the series ,we conclude as follows */

$\boxed { \pink { a_{n} = (3n)^{2} , \: where \: n\in N }}$

$i)First \:term (a_{1} )\\ = (3\times 1)^{2}\\ = 3^{2}\\ = 9$

$ii)Second \:term (a_{2} ) \\= (3\times 2)^{2} \\ = 6^{2}\\ = 36$

$iii)Third \:term (a_{3} ) \\= (3\times 3)^{2} \\ = 9^{2}\\ = 81$

$iv) Fourth \:term (a_{4} ) \\= (3\times 4)^{2} \\ = 12^{2}\\ = 144$

$\red{ v)Next \:term (a_{5} ) }\\= (3\times 5)^{2} \\ = 15^{2}\\ \green {= 225}$

Therefore.,

$\red{ Next \:term }\green {= 225}$

•••♪