Question

Which term of the arithmetic progression $8,14,20,26,....$ will be $72$ more than its $41$st term.

Answer 1

Given sequence is $8,14,20,26,...$

The first term is $a=8$

The common difference is $d=14-8=6$

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

Let the $m^{th}$ term be $72$ more than the ${41}^{st}$ term

Therefore, $72+m^{th}term={41}^{st}term$

$⟹72+a+(m-1)d=a+(41-1)d$

$⟹72+(m-1)6=\left(40\right)6$

$⟹6m-6=240+72$

$⟹6m=312+6$

$⟹6m=318$

$⟹m=\frac{318}{6}=53$

Therefore, the ${53}^{rd}$ term is $72$ more than the ${41}^{st}$ term