Question

The ${17}^{th}$ term of an A.P. is 5 more than twice its $8^{th}$ term. If the ${11}^{th}$ term of the A.P. is 43 find the $n^{th}$ term.

Answer 1

$a_{17}=2a_8+5a+16d=2(a+7d)+5a+16d=2a+14d)+5-a+2d=5⟶(i)$ $a_{11}=43a+10d=43⟶\left(ii\right)$

Adding $\left(i\right)$ and $\left(ii\right)$.

$-a+2d=5$

$a+10d=43$

$12d=48=4a=3$

$a_n=a+(n-1)d$

$a_n=3+(n-1)4$

$a_n=4n-1$