Let a and b be two distinct positive real numbers. Let 11th term of a GP, whose first term is a and third term is b, is equal to pth term of another GP, whose first term is a and fifth term is b. Then p is equal to
(3) 21
GP 1: ar2=b r2= ba t11=ar10=a ( ba )5
GP 2: aR4=b R= ( ba )1/4 Tp=aRp-1=a ( ba ) p-14
t11=Tp, ∵ a b ba 1
5= p-14 p=21
Answer: (3)
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