Question

For a positive number 'b', the value of 'b' for which the numbers 3^x + 3^-x , b,9^x+ 9-x are in
A.P. can be :
(A) 1
(B) 2
(C) 3
(D) 5​

Answer 1

Answer this question

d) 5

Answer 2

None of these.

Concept:

The differences between every two successive words are the same in an arithmetic progression (AP).  It's a sequence in which each phrase, except the first, is obtained by multiplying the previous term by a fixed integer.

Given:

The numbers $3^x+3^{-x}, b, 9^x+9^{-x}$ are in A.P.

Find:

The value of $b$.

Solution:

In an A.P, the common difference between any two consecutive numbers is same.

$b-(3^x+3^{-x}) = 9^x+9^{-x}-b\\\\2b = 9^x+9^{-x} + 3^x+3^{-x}\\\\2b = 3^{2x}+3^{-2x}+3^x+3^{-x}\\\\2b = 3^x(3^x+1)+3^{-x}(3^{-x}+1)\\\\2b = 3^x(3^x+1)+3^{-x}(\frac{1+3^{x}}{3^{x}}) \\\\2b = 3^x(3^x+1)+3^{-2x}(1+3^{x})\\\\2b = (3^x+1)(3^x+3^{-2x})$

$b = \frac{ (3^x+1)(3^x+3^{-2x})}{2}$

The value of $b$ depends on the value of $x$.

Hence, the value of $b$ is $b = \frac{ (3^x+1)(3^x+3^{-2x})}{2}$. Option(A) is incorrect.

Hence, the value of $b$ is $b = \frac{ (3^x+1)(3^x+3^{-2x})}{2}$. Option(B) is incorrect.

Hence, the value of $b$ is $b = \frac{ (3^x+1)(3^x+3^{-2x})}{2}$. Option(C) is incorrect.

Hence, the value of $b$ is $b = \frac{ (3^x+1)(3^x+3^{-2x})}{2}$. Option(D) is incorrect.