Since, the first 11 terms are Ap,d=2 ∴a11=a+10d=a+20 The middle term of AP is T6=a+5d=a+10 For the next 11 terms in GP r=2 ∴ The middle term of GP is b(2)5 where, b is the first term of a GP which is the last term of AP b(2)5=(a+20)32 According to the given condition, ⇒a+10=(a+20)32 ⇒32a=10−640 a=−
630
31
∴ Middle term of entire sequence is 11th term ∴T11=