Let first term and common difference of an AP are a and d. ∴T3=a+2d T7=a+6d and T12=a+11d But T3,T7 and T12 are consecutive terms of an GP. ∴(a+6d)2=(a+2d)(a+11d)
a2+36d2+12ad=a2+13ad+22d2
ad=14d2 ⇒a=14d ∴ Terms are 14d+2d,14d+6d,14d+11d ie., 16d,20d,25d ∴ Common ratio =