Question 277454
the sum of the first six terms of a geometric progression is nine time the sum of the first three terms, where neither the first nor the common ratio is equal to zero. find the common ratio.
<pre><font size = 4 color = "indigo"><b>

Let the first term be {{{a}}} and the common ratio be {{{r}}}

Sum of first three terms = {{{a+ar+ar^2}}}

Let this sum of the first three terms be {{{X}}}

Sum of first six terms = {{{a+ar+ar^2+ar^3+ar^4+ar^5}}}

Sum of first six terms = {{{X+ar^3+ar^4+ar^5=X+r^3(a+ar+ar^2)=X+r^3X}}}

We are given that {{{X+r^3X=9X}}}.  Divide through by X and we have

{{{1+r^3=9}}}

{{{r^3=8}}}

{{{r=root(3,8)}}}

{{{r=2}}}

Check: Suppose the first term is a. Then the first six terms are

a,2a,4a,8a,16a,32a

The sum of the first three terms is a+2a+4a = 7a

The sum of the first 6 terms is a+2a+4a+8a+16a+32a=63a which is 9 times 7a.

So the answer is r=2.

Edwin</pre>