Question 1186218
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Let the first term be a and the common ratio be r.<br>
(1) The infinite sum is {{{a/(1-r)}}}<br>
{{{a/(1-r)=108}}}
{{{a=108(1-r)}}}<br>
The sum of the first three terms is {{{a+ar+ar^2}}}<br>
{{{a+ar+ar^2=112}}}
{{{a(1+r+r^2)=112}}}
{{{(108(1-r))(1+r+r^2)=112}}}
{{{108(1-r^3)=112}}}
{{{1-r^3=112/108=28/27}}}
{{{r^3=-1/27}}}
{{{r=-1/3}}}<br>
{{{a=108(1-r) = 108(1+1/3)=144}}}<br>
ANSWER: The first term is a = 144.<br>
CHECK:<br>
infinite sum: {{{a/(1-r)=144/(1+1/3) = 144/(4/3) = 108}}}<br>
sum of first three terms: {{{144+(-48)+16 = 112}}}<br>