Question 1104515
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The sum of an infinite geometric series with first term a and common ratio r (|r|<1) is<br>
{{{S = a/(1-r)}}}<br>
We are told that
{{{1/(1-r) = 1.85}}}
{{{1.85-1.85r = 1}}}
{{{1.85r = .85}}}
{{{r = .85/1.85 = 17/37}}}<br>
The series whose sum we are to find has first term r and common ratio r^2; the sum of that infinite series is
{{{r/(1-r^2) = (17/37)/(1-(17/37)^2) = (17/37)/(1-289/1369) = (17/37)/((1369-289)/1369) = (17/37)(1369/1080) = 17*37/1080 = 629/1080}}}