JEE Main 2024 — Sequences And Series Question with Solution

To determine the least value of $$\mathrm{K}$$ for which the terms $$4^{1+x}+4^{1-x},\frac{\mathrm{K}}{2},16^x+16^{-x}$$ form an arithmetic progression (A.P.), we need to establish the relationship among these terms in an A.P.

For three numbers to be in an arithmetic progression, the middle term must be the average of the other two terms. Therefore, we can write:

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

First, simplify each term individually:

1. Consider $$4^{1+x}+4^{1-x}$$:

$$4^{1+x}=4\cdot 4^x=4\cdot (2^2{)}^x=4\cdot 2^{2x}=4\cdot 2^{2x}$$

and

$$4^{1-x}=4\cdot 4^{-x}=4\cdot (2^2{)}^{-x}=4\cdot 2^{-2x}=4\cdot 2^{-2x}$$

Thus,

$$4^{1+x}+4^{1-x}=4\cdot 2^{2x}+4\cdot 2^{-2x}=4(2^{2x}+2^{-2x})$$

2. Consider $$16^x+16^{-x}$$:

$$16^x=(2^4{)}^x=2^{4x}$$

and

$$16^{-x}=(2^4{)}^{-x}=2^{-4x}$$

Thus,

$$16^x+16^{-x}=2^{4x}+2^{-4x}$$

3. Combine the terms and set up the equation:

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

Multiply both sides by 2:

$$4(2^{2x}+2^{-2x})+2^{4x}+2^{-4x}=K$$

To find the least value of $$\mathrm{K}$$, let's assume $$x=0$$ (since $$x$$ can range over non-negative values):

For $$x=0$$:

$$4(2^{2\cdot 0}+2^{-2\cdot 0})+2^{4\cdot 0}+2^{-4\cdot 0}$$

This simplifies to:

$$4(2^0+2^0)+2^0+2^0$$

$$=4(1+1)+1+1$$

$$=4\cdot 2+1+1$$

$$=8+1+1$$

$$=10$$

Therefore, the least value of $$\mathrm{K}$$ that ensures the values form an arithmetic progression is $10$. Hence, the correct option is:

Option A: 10