Question

If the 10th term of an AP is 52 and 17th term is 20 more than its 13th term, find the AP.

Answer 1

given, $a_{10}=52$
$a+9d=52$ ...........(1)
also, $a_{17}=20+a_{13}$
$a+16d=20+a+12d$
$16d-12d=20$
$4d=20$
$d=5$
putting the value of d in eq. (1), we get,
$a+9\left(5\right)=52$
$a+45=52$
$a=7$
hence, the required AP is 7,12,17,.........