For x \geqslant 0 , the least value of K , for which 4^1+x+4^1-x, \fracK2, 16^x+16^-x are three consecutive terms of an A.P., is equal to :

Sequences and Series PYQ — JEE Main 2024 · 5 Apr · Shift 2 · Q46 with Solution

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Correct answer: A

Let $A=4^{1+x}+4^{1-x},\qquad B=\frac K2,\qquad C=16^x+16^{-x}.$ Since these are three consecutive terms of an A.P., the middle term is the average of the other two: $2B=A+C.$ Because $B=\dfrac K2$ , we get $K=A+C.$ So we need the least value of $K=4^{1+x}+4^{1-x}+16^x+16^{-x},\qquad x\ge 0.$
Rewrite everything in terms of $t=4^x$ . Since $x\ge 0$ , we have $t=4^x\ge 1$ . Also, $4^{1+x}=4\cdot 4^x=4t,\qquad 4^{1-x}=\frac{4}{4^x}=\frac{4}{t},$ $16^x=(4^2)^x=4^{2x}=t^2,\qquad 16^{-x}=t^{-2}=\frac{1}{t^2}.$ Hence $K=t^2+4t+\frac{4}{t}+\frac{1}{t^2}.$
Observe that

=\left(t+\frac{1}{t}\right)^2+4\left(t+\frac{1}{t}\right)-2.$$ Let $$u=t+\frac{1}{t}.$$ Since $t\ge 1$, we have $u\ge 2$. Therefore, $$K=u^2+4u-2.$$ 4. Now minimize this for $u\ge 2$. The function $$f(u)=u^2+4u-2$$ is increasing for $u\ge 2$ because $$f'(u)=2u+4>0.$$ So the minimum occurs at the smallest possible value of $u$, namely $u=2$. Thus, $$K_{\min}=2^2+4(2)-2=4+8-2=10.$$ 5. This occurs when $$t+\frac{1}{t}=2 \implies t=1,$$ so $4^x=1\implies x=0$, which is allowed. 6. Therefore the least value of $K$ is $$\boxed{10}.$$ Checking options: Option A is correct.