UPSC NDA I 2025 Mathematics Paper

Let the 5th, 7th, and 13th terms of the AP be T5,T7, and T13 respectively.
‌⇒T5=a+4d
‌⇒T7=a+6d
‌⇒T13=a+12d
Since the terms are in GP:
‌⇒T72=T5×T13
‌⇒(a+6d)2=(a+4d)×(a+12d)
Expanding both sides:
‌⇒(a+6d)2=a2+12ad+48d2
‌⇒a2+12ad+36d2=a2+12ad+48d2
Cancel out common terms:
‌⇒36d2=48d2
‌⇒−12d2=0
Dividing through by d2 (assuming d≠0 ):
⇒a=−3d