Question 1181525
<br>
Presumably the first statement is supposed to be....<br>
the sum of the {{{cross(first)}}} {{{cross(term)}}} first and second terms of an exponential sequence is 135.<br>
Then the problem has a "nice" solution.<br>
Let a and r be the first term and common ratio.  Then<br>
{{{a+ar=135}}}
{{{ar^2+ar^3 = r^2(a+ar) = r^2(135) = 60}}}<br>
{{{r^2 = 60/135 = 4/9}}}<br>
Since we are told r is positive, r is 2/3.  Then<br>
{{{a+ar = a+(2/3)a = (5/3)a = 135}}}
{{{a = (3/5)(135) = 81}}}<br>
The infinite sum is first term divided by (1 minus the common ratio):<br>
{{{a/(1-r) = 81/(1-2/3) = 81/(1/3) = 81*3 = 243}}}<br>
ANSWER: 243<br>