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

Hint:

For solving geometric progressions, use the sum and product formulas systematically, and simplify using algebraic techniques like substitution and factorization.

Answer 1

The Correct Option is C

To solve the problem, we need to find the sum of the common ratios of all possible geometric progressions (GPs) with four positive consecutive terms, given that their sum is 126 and the product is 1296.

Let's denote the four consecutive terms of the GP as \(a, ar, ar^2, ar^3\), where \(a\) is the first term and \(r\) is the common ratio.
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