Question

The sum of the series, $S=\frac{7}{1^2\cdot 6^2}+\frac{17}{6^2\cdot 11^2}+\frac{27}{11^2\cdot 16^2}+\frac{37}{16^2\cdot 21^2}+\dots \dots +\infty$ terms is equal to $\frac{p}{q}$ (where $p,q\in N$ and $p,q$ are coprime). If arithmetic mean of two positive numbers be ( $p$ $+q)$ and their geometric mean $G$ and harmonic mean $H$ satisfy the relation $G^2+6H=32$ then find square root of product of both the numbers.

Answer 1

Answer: 4

$\Theta S=\frac{7}{1^2\cdot 6^2}+\frac{17}{6^2\cdot 11^2}+\frac{27}{11^2\cdot 16^2}+\frac{37}{16^2\cdot 21^2}+\dots \dots$
$\Rightarrow 5S=\frac{(6-1)(6+1)}{1^2\cdot 6^2}+\frac{(11-6)(11+6)}{6^2\cdot 11^2}+\frac{(16-11)(16+11)}{11^2\cdot 16^2}+\frac{(21-16)(21+16)}{16^2\cdot 21^2}+\dots \dots .$.
$=\frac{6^2-1^2}{1^2\cdot 6^2}+\frac{11^2-6^2}{6^2\cdot 11^2}+\frac{16^2-11^2}{11^2\cdot 16^2}+\frac{21^2-16^2}{16^2\cdot 21^2}+\dots \dots ..$
$=\left(1-\frac{1}{6^2}\right)+\left(\frac{1}{6^2}-\frac{1}{11^2}\right)+\left(\frac{1}{11^2}-\frac{1}{16^2}\right)+\left(\frac{1}{16^2}-\frac{1}{21^2}\right)+\dots \dots \dots \infty \text{ terms }$
$\Rightarrow 5S=1\Rightarrow S=\frac{1}{5}=\frac{p}{q}\therefore p+q=1+5=6$
Let two positive real number be $a$ and $b$
$\therefore \frac{a+b}{2}=6\Rightarrow a+b=12$
$\Theta G^2+6H=32\Rightarrow ab+6\cdot \frac{2ab}{a+b}=32\Rightarrow ab+ab=32\Rightarrow ab=16$
$\therefore \sqrt{ab}=4$