JEE Main | 2014Binomial TheoremHard
Question
Three positive numbers form an increasing G.P. If the middle term in this G.P. is doubled, the new numbers are in A.P. Then the common ratio of the G.P. is
Options
A.2 - √3
B.2 + √3
C.√2 + √3
D.3 + √2
Solution
a, ar, ar2 → G.P.
a, 2ar, ar2 → A.P.
2 × 2ar = a + ar2
4r = 1 + r2
⇒ r2 - 4r + 1 = 0
r =
= 2 ± √ 3

r = 2 - √3 is rejected
∵ (r > 1)
G.P. is increasing.
a, 2ar, ar2 → A.P.
2 × 2ar = a + ar2
4r = 1 + r2
⇒ r2 - 4r + 1 = 0
r =
r = 2 - √3 is rejected
∵ (r > 1)
G.P. is increasing.
Create a free account to view solution
View Solution FreeMore Binomial Theorem Questions
The expression [x + (x3 − 1)1/2]5 + [x − (x3 − 1)1/2]5 is a polynomial of degree -...The value of the expression 47C4 + 52 - jC3 is equal to :...If (1 + by)n = (1 + 8y + 24 y2 + ....) then the value of b and n are respectively-...In the expansion of , n ∈ N, if the sum of the coefficients of x5 and x10 is 0, then n is :..., then =...