Question

Column II gives $n^{th}$ term for AP given in column I. Match them correctly.

Answer 1

By putting $n=1$ in general terms of all APs given, we get
$13-3n=13-3\left(1\right)=10$
$9-5n=9-5\left(1\right)=4$
$3+4n=3+4\left(1\right)=7$
$17n+102=17\left(1\right)+102=119$
Now, the series can be easily matched.