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

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$\frac{9}{2}$

answer is D.

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Detailed Solution

Given the sequence a, ar, ar², ar³ (where a, r > 0):

a⁴r⁶ = 1296
a²r³ = 36
Dividing the equations, we get:
a^2r^3 = 36 ⇒ a = 6 / r^3/2
The sum of the sequence is:
a + ar + ar² + ar³ = 126
Substituting a = 6 / r^3/2:
(1 / r^3/2) + (r / r^3/2) + (r² / r^3/2) + (r³ / r^3/2) = 126 / 6
Simplifying:
r^-3/2 + r^-1/2 + r^1/2 + r^3/2 = 21
Grouping terms:
(r^-3/2 + r^3/2) + (r^1/2 + r^-1/2) = 21
Let r^1/2 + r^-1/2 = A:
r^-3/2 + r^3/2 + 3A = A³
Substituting A³ - 3A + A = 21:
A³ - 2A = 21
Solving A = 3:
√r + 1 / √r = 3
Squaring both sides:
r + 1 + (1 / r) = 9 ⇒ r² + 2r + 1 = 9r
Simplifying:
r² - 7r + 1 = 0
Solving r, we find the correct value of r = 7. Hence, the correct option is (1) 7.

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