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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Sequence and Series
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In a GP, the first term is $312\large\frac{1}{2}$, the common ratio is $\large\frac{1}{2}$. Find the sum of the series to $\infty$ terms.

$\begin{array}{1 1} 625 \\ {312}\;\frac{1}{2} \\ 1250 \\ \infty \end{array}$

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1 Answer

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answer is 625
Hint:$ S_\infty=\large\frac{a}{1-r}$
Answer: In a GP,$S _\infty=\large\frac{a}{1-r}$
a=${312}\;\large\frac{1}{2}$ ,r=$\large\frac{1}{2}$
$S_\infty=\large\frac{a}{1-r}$
$\large\frac{312\;1/2}{1-\Large\frac{1}{2}}$=312.5*2
=625
answered Dec 30, 2013 by yamini.v
edited Dec 30, 2013 by sreemathi.v
 

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