Question

The sum of the series, $S=\frac{7}{1^2 \cdot 6^2}+\frac{17}{6^2 \cdot 11^2}+\frac{27}{11^2 \cdot 16^2}+\frac{37}{16^2 \cdot 21^2}+\ldots \ldots+\infty$ 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 $G ^2+6 H =32$ then find square root of product of both the numbers.

Answer 1

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