Given, the 2nd term of an AP is 13.
⇒a2=13
⇒a+(2−1)d=13 [∵an=a+(n−1)d]
and the 5th term of an AP is 25
⇒a5=25
⇒a+(5−1)d=25
⇒a+d=13...(i)
and a+4d=25...(ii)
On subtracting eq.(i) from eq. (ii), we get,
3d = 25 - 13 = 12
⇒ d = 4
From eq. (i), a = 13 - 4 = 9
∴ a7=a+(7−1)d=9+6× 4=33