For x ? 0 , the least value of K , for which 4 1 + x + 4 1 - x , ? K 2 , ? 16 x + 16 - x are three consecutive terms of an A.P., is equal to: [2024]

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
For $x \geq 0,$ the least value of $K,$ for which $4^{1 + x} + 4^{1 - x}, \frac{K}{2}, 16^{x} + 16^{- x}$ are three consecutive terms of an A.P., is equal to: [2024]

1 4

2 10

3 8

4 16

Ans.
(2)

We have, $4^{1 + x} + 4^{1 - x}, \frac{K}{2}, 16^{x} + 16^{- x}$ are in A.P.

$\Rightarrow 2 \frac{K}{2} = (4^{1 + x} + 4^{1 - x}) + (16^{x} + 16^{- x})$

$\Rightarrow K = 4 \cdot 4^{x} + \frac{4}{4^{x}} + 4^{2 x} + \frac{1}{4^{2 x}}$

$= 4 (4^{x} + \frac{1}{4^{x}}) + (4^{2 x} + \frac{1}{4^{2 x}})$ $\geq 4 \cdot 2 + 2$ $[∵ A . M . \geq G . M .]$

$= 10 \Rightarrow K \geq 10$

So, least value of $K$ is 10

Want Us to Email you About Special Offers & Updates?