# In a GP, given that the first term is $312 ½$ and the common ratio is $1/2$ find the sum of the terms of the series to infinity.

$(a)\;625\qquad(b)\;1250\qquad(c)\;312 \frac{1}{2}\qquad(d)\;\infty$

it is provided that first term a=625/2

r=1/2

sum to infinity= a/(1-r)

=625/2(1-1/2)

=1250/2 ===625

so a is the correct one

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