Question

If $x=\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+....,$ $y=\frac{1}{1^2}+\frac{3}{2^2}+\frac{1}{3^2}+\frac{3}{4^2}+..$ and $z=\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{3^2}-\frac{1}{4^2}+....,$ then

• A. $x,y,z$ are in A.P.
• B. $\frac{y}{6},\frac{x}{3},\frac{z}{1}$ are in A.P.
• C. $\frac{y}{6},\frac{x}{3},\frac{z}{2}$ are in A.P.
• D. $6y,3x,2z$ are in H.P.

Answer 1

The correct option is C $\frac{y}{6},\frac{x}{3},\frac{z}{2}$ are in A.P.

$y-x=[\frac{1}{1^2}+\frac{3}{2^2}+\frac{1}{3^2}+\frac{3}{4^2}+...]-[\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+...]$

$=3[\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...]......\left(1\right)$

Similarly,

$x-z=[\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+....]-[\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{3^2}-\frac{1}{4^2}+....]$

$=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}........\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$,

$y-x=3(x-z)⟹y+3z=4x$

$\Rightarrow \left(\frac{y}{6}\right)+\left(\frac{z}{2}\right)=2\left(\frac{x}{3}\right)$ (dividing both sides by $6$)

Therefore $\frac{y}{6},\frac{x}{3}and\frac{z}{2}$ are in A.P