If the 2nd term of an AP is 13 and

(b) Given, a2 = 13 and a5 = 25

$\Rightarrow \quad a+(2-1) d=13 \quad\left[\because a_{n}=a+(n-1) d\right]$

and $\quad a+(5-1) d=25$

$\Rightarrow \quad a+d=13 \quad \ldots$ (i)

and  $a+4 d=25$ ...(ii)

On subtracting Eq. (i) from Eq. (ii), we get

$3 d=25-13=12 \Rightarrow d=4$

From Eq. (i), $a=13-4=9$

$\therefore \quad a_{7}=a+(7-1) d=9+6 \times 4=33$