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Sequence and Series
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.
$\begin{array}{1 1} 625 \\ {312}\;\frac{1}{2} \\ 1250 \\ \infty \end{array}$
jeemain
math
class11
ch9
sequences-and-series
easy
geometric-progression
q1
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asked
Dec 30, 2013
by
yamini.v
edited
Apr 1, 2016
by
pady_1
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1 Answer
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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