Question

If the sum and product of four positive consecutive terms of a GP, are 126 and 1296, respectively, then the sum of common ratios of all such GPs is

• A. $\frac{9}{2}$
• B. 3
• C. 7
• D. 14

Answer 1

Answer: (C)

7

Answer 2

a,ar,ar2,ar3(a,r>0)
a4r6=1296
a2r3=36
a=r3/26​
a+ar+ar2+ar3=126
r3/21​+r3/2r​+r3/2r2​+r3/2r3​=6126​=21
(r−3/2+r3/2)+(r1/2+r−1/2)=21
r1/2+r−1/2=A
r−3/2+r3/2+3A=A3
A3−3A+A=21
A3−2A=21
A=3
r​+r​1​=3
r+1=3r​
r2+2r+1=9r
r2−7r+1=0