Question

Which term of the A.P. $3,15,27,39,....$ will be $120$ more than its $21$st term?

Answer 1

Given sequence is $3,15,27,39,...$

The first term is $a=3$

The common difference is $d=15-3=12$

we know that the $n^{th}$ term of the arithmetic progression is given by $a+(n-1)d$

Let the $m^{th}$ term be $120$ more than the ${21}^{st}$ term

Therefore, $120+m^{th}term={21}^{st}term$

$⟹120+a+(m-1)d=a+(21-1)d$

$⟹120+(m-1)12=\left(20\right)12$

$⟹12m-12=240+120$

$⟹12m=360+12$

$⟹12m=372$

$⟹m=\frac{372}{12}=31$

Therefore, the ${31}^{st}$ term is $120$ more than the ${21}^{st}$ term