Question

If the ${10}^{th}$ term of an A.P. is 52 and the ${17}^{th}$ term is 20 more than the ${13}^{th}$ term, find the A.P.

Answer 1

Given $a_{10}=52$

$a_n=a+(n-1)d52=a+9d\left(1\right)$

also, $a_{17}=20+a_{13}a+16d=20+a+12d4d=20d=5\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$

$52=a+9\times 5a=7$

The AP is: $7,12,17,22,...$