Question 1090598
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The sum to infinity of a geometric sequence is 27/2 while the sum of the first three terms is 13. Find the sum of the first 5 terms.

Thank you for your help.
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<pre>
We have 

S = {{{a + aq + aq^2 + aq^3 + aq^4 + aq^5 + ellipsis}}} = {{{(a + aq + aq^2)}}} + {{{(aq^3 + aq^4 + aq^5 + ellipsis)}}} = {{{(a + aq + aq^2)}}} + {{{q^3*(a + aq + aq^2 + ellipsis)}}}   (1)


Notice that the infinite sum in parentheses of the right-most side is S, again.   // <U>It is the KEY IDEA #1</U>.


Based on given info, replace in (1)  S  by {{{27/2}}}  and replace the sum of the first three terms by 13. You will get

{{{27/2}}} = {{{13 + q^3*S}}},    or, which is the same

{{{1/2}}} = {{{q^3*S}}}.    (2).


In (2),  replace S by  {{{27/2}}}  as it is given.  // <U>It is the KEY IDEA #2</U>.


You will get   {{{1/2}}} = {{{q^3*(27/2)}}},   which implies   {{{q^3}}} = {{{1/27}}}.


Hence,  q = {{{1/3}}}.   Thus we just found the common ratio of the progression;  it is  q = {{{1/3}}}. 


Now we are at the finish line.


Similar to (1), we have

S = {{{a + aq + aq^2 + aq^3 + aq^4 + aq^5 + aq^6 + aq^7 + ellipsis}}} = {{{(a + aq + aq^2 + aq^3 + aq^4)}}} + {{{(aq^5 + aq^6 + aq^7 + ellipsis)}}} = {{{(a + aq + aq^2 + aq^3 + a^4)}}} + {{{q^5*(a + aq + aq^2 + ellipsis)}}}   (3)


You can re-write it as


S = {{{S[5]}}} + {{{q^5*S}}},     (4)      // <U>It is the KEY IDEA #3</U>.


where {{{S[5]}}}  is the sum of the first 5 terms, which is under the question.

Now from (4)

{{{S[5]}}} = {{{S - q^5*S}}} = {{{S*(1-q^5)}}} = {{{(27/2)*(1 - (1/3)^5)}}} = {{{(27/2)*(1-1/243)}}} = {{{(27/2)*(242/243)}}} = {{{(27/243)*(242/2)}}} = {{{(1/9)*121}}} = {{{121/9}}} = {{{13}}} {{{4/9}}}.
</pre>

<U>Answer</U>.  The sum of the first 5 terms of the given GP is  {{{13}}} {{{4/9}}}.



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