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Let the greatest common divisor of $m, n$ be 1 . If $\frac{1}{1 \cdot 7}+\frac{1}{7 \cdot 13}+\frac{1}{13 \cdot 19}+\ldots . .$. upto 20 terms $=\frac{m}{n}$, then $5 m+2 n=$
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342
(c) We have,
$\begin{aligned}
& \frac{m}{n}=\frac{1}{1 \cdot 7}+\frac{1}{7 \cdot 13}+\frac{1}{13 \cdot 19}+\ldots+\frac{1}{[1+(20-1) 6][7+(20-1) 6]} \\ & =\frac{1}{1 \cdot 7}+\frac{1}{7 \cdot 13}+\frac{1}{13 \cdot 19}+\ldots+\frac{1}{115 \times 121}
\end{aligned}$
$\begin{aligned} & =\frac{1}{6}\left[\frac{6}{1 \cdot 7}+\frac{6}{7 \cdot 13}+\frac{6}{13 \cdot 19}+\ldots+\frac{6}{115 \times 121}\right] \\ & =\frac{1}{6}\left[\frac{7-1}{1 \times 7}+\frac{13-7}{7 \times 13}+\frac{19-13}{13 \times 19}+\ldots+\frac{121-115}{115 \times 121}\right] \\ & =\frac{1}{6}\left[\left(\frac{1}{1}-\frac{1}{7}\right)+\left(\frac{1}{7}-\frac{1}{13}\right)+\left(\frac{1}{13}-\frac{1}{19}\right)+\ldots+\left(\frac{1}{115}-\frac{1}{121}\right)\right] \\ & =\frac{1}{6}\left[1-\frac{1}{121}\right]=\frac{1}{6} \times \frac{120}{121}=\frac{20}{121} \quad \therefore \quad m=20, n=121 \\ & \therefore \quad 5 m+2 n=5 \times 20+2 \times 121=100+242=342\end{aligned}$
$\begin{aligned}
& \frac{m}{n}=\frac{1}{1 \cdot 7}+\frac{1}{7 \cdot 13}+\frac{1}{13 \cdot 19}+\ldots+\frac{1}{[1+(20-1) 6][7+(20-1) 6]} \\ & =\frac{1}{1 \cdot 7}+\frac{1}{7 \cdot 13}+\frac{1}{13 \cdot 19}+\ldots+\frac{1}{115 \times 121}
\end{aligned}$
$\begin{aligned} & =\frac{1}{6}\left[\frac{6}{1 \cdot 7}+\frac{6}{7 \cdot 13}+\frac{6}{13 \cdot 19}+\ldots+\frac{6}{115 \times 121}\right] \\ & =\frac{1}{6}\left[\frac{7-1}{1 \times 7}+\frac{13-7}{7 \times 13}+\frac{19-13}{13 \times 19}+\ldots+\frac{121-115}{115 \times 121}\right] \\ & =\frac{1}{6}\left[\left(\frac{1}{1}-\frac{1}{7}\right)+\left(\frac{1}{7}-\frac{1}{13}\right)+\left(\frac{1}{13}-\frac{1}{19}\right)+\ldots+\left(\frac{1}{115}-\frac{1}{121}\right)\right] \\ & =\frac{1}{6}\left[1-\frac{1}{121}\right]=\frac{1}{6} \times \frac{120}{121}=\frac{20}{121} \quad \therefore \quad m=20, n=121 \\ & \therefore \quad 5 m+2 n=5 \times 20+2 \times 121=100+242=342\end{aligned}$
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