Question 1122514
<br>
Let the GP be<br>
a, ar, ar^2, ar^3, ...<br>
Then the GP with the terms squared is<br>
a^2, a^2r^2, a^2r^4, a^2r^6, ...<br>
The infinite sum of the GP is<br>
(1) {{{a/(1-r) = 15}}}<br>
The infinite sum of the squared GP is<br>
(2) {{{a^2/(1-r^2) = 45}}}<br>
Dividing (2) by (1) gives us<br>
(3) {{{a/(1+r) = 3}}}<br>
Then (1) and (3) give us two linear equations in a and r that we can solve:<br>
{{{a = 15-15r}}}
{{{a = 3+3r}}}
{{{15-15r = 3+3r}}}
{{{12 = 18r}}}
(4) {{{r = 2/3}}}<br>
Then substituting (4) in (1) gives us a=5.<br>
The GP has first term 5 and common ratio 2/3:<br>
5, 10/3, 20/9, ...<br>
The squared GP is<br>
25, 100/9 400/81, ...<br>
CHECK:<br>
The sum of the GP is<br>
{{{5/(1-(2/3)) = 5/(1/3) = 15)}}}<br>
The sum of the squared GP is<br>
{{{25/(1-(4/9)) = 25/(5/9) = 45}}}<br>