The 3rd and 6th term of a G.P. is 12 and 96 respectively. If the sum of all terms is 1533, find the number of terms in the G.P.
9
Given, a3=ar2=12 and a6=ar5=96.
⇒ar5ar2=9612=8
⇒r3=8⇒r=2
∴ar2=12⇒a×22=12
⇒a=3
Let the number of terms in the GP be n.
Sum to n terms of a GP of first term a and common ratio r(≠1) is given by
Sn=a(rn−1)r−1.
⇒3(2n−1)2−1=1533
⇒2n−1=511
⇒2n=512
i.e., 2n=29
∴n=9