Question

Which term of the A.P. 8,14,20,26.... will be 72 more than its $41st$ term?

Answer 1

AP $:8,14,20,\mathrm{26...}$

$a=8$

$d=14-8=6$

Let $T_n=T_{41}+72$

$a+(n-1)d=a+40d+72$

$\Rightarrow a+(n-1)d-a-40d=72$

$\Rightarrow d(n-1-40)=72$

$\Rightarrow 6(n-41)=72$

$\Rightarrow n-41=12$

$\Rightarrow n=12+41$

$\therefore n=53$

Required term = 53

${53}^{th}$ term of AP will be 72 more than its ${41}^{th}$ term