Question 898542
In a geometric series, the sum of the first three terms is 304, 
<pre>
{{{a[1]+a[1]r+a[1]r^2}}}{{{""=""}}}{{{304}}}

(1)   {{{a[1](1+r+r^2)}}}{{{""=""}}}{{{304}}}
</pre>
and the sum of the first six terms is 1330. 
<pre>
So the sum of the 3rd, 4th and 5th terms only is 1330-304 = 1026

{{{a[1]r^3+a[1]r^4+a[1]r^5}}}{{{""=""}}}{{{1026}}}

(2)   {{{a[1]r^3(1+r+r^2)}}}{{{""=""}}}{{{1026}}}

Dividing equals by equals, equation (2) by equation (1)

    {{{(a[1]r^3(1+r+r^2))/(a[1](1+r+r^2))}}}{{{""=""}}}{{{1026/304}}}

    {{{(cross(a[1])r^3cross((1+r+r^2)))/(cross(a[1])cross((1+r+r^2)))}}}{{{""=""}}}{{{27/8}}}

    {{{r^3}}}{{{""=""}}}{{{27/8}}}

Taking cube roots of both sides

    {{{r}}}{{{""=""}}}{{{3/2}}}

Substitute in equation (1)

(1)   {{{a[1](1+r+r^2)}}}{{{""=""}}}{{{304}}}

      {{{a[1](1+3/2+(3/2)^2)}}}{{{""=""}}}{{{304}}}

      {{{a[1](1+3/2+9/4)}}}{{{""=""}}}{{{304}}}

Multiply the terms in the parentheses and the right side by 4
to clear of fractions:

      {{{a[1](4+6+9)}}}{{{""=""}}}{{{1216}}}

      {{{a[1](19)}}}{{{""=""}}}{{{1216}}}

Divide both sides by 19

      {{{a[1]}}}{{{""=""}}}{{{1216/19}}}

      {{{a[1]}}}{{{""=""}}}{{{64}}}
</pre>
Find the sum of the first seven terms.
<pre>
{{{S[n]}}}{{{""=""}}}{{{(a[1](r^n-1))/(r-1)}}}

{{{S[7]}}}{{{""=""}}}{{{(64((3/2)^7-1))/(3/2-1)}}}

{{{S[7]}}}{{{""=""}}}{{{(64(2187/128-1))/(3/2-1)}}}

Multiply the top out, the 16 goes into the 128 2 times,
and write the 1 on the bottom as {{{2/2}}}

{{{S[7]}}}{{{""=""}}}{{{(2187/2-64)/(3/2-2/2)}}}

{{{S[7]}}}{{{""=""}}}{{{(2187/2-64)/(1/2)}}}

Multiply top and bottom by 2

{{{S[7]}}}{{{""=""}}}{{{2(2187/2-64)/(2*expr(1/2))}}}
 
{{{S[7]}}}{{{""=""}}}{{{(2187-128)/1}}}

{{{S[7]}}}{{{""=""}}}{{{2059}}}

Edwin</pre>