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 =
= 2 ± √ 3r = 2 - √3 is rejected
∵ (r > 1)
G.P. is increasing.
Create a free account to view solution
View Solution FreeMore Binomial Theorem Questions
The total number of distinct terms in the expansion of, (x + a)100 + (x - a)100 after simplification is :...If x + y = 1, then equals -...Let f(n) = 10n + 3.4n +2 + 5, n ∈ N. The greatest value of the integer which divides f(n) for all n is :...The coefficient of x5 in the expansion of (x2 − x − 2)5 is -...Sum of the infinite series + ..... to ∞...