Question 1206544
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Find the sum of the first 8 terms of a geometric series if the first term is 10 and the common ratio is 3.
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<pre>
Use the formula for the sum of the first n terms of a GP 

    {{{S[n]}}} = {{{a[1]*((r^n-1)/(r-1))}}} = {{{10*((3^8-1)/(3-1))}}} = {{{10*(6560/2)}}} = 10*3280 = 32800.    <U>ANSWER</U>
</pre>

Solved.


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On geometric progressions, &nbsp;see introductory lessons

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/Sequences-and-series/Geometric-progressions.lesson>Geometric progressions</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/Sequences-and-series/The-proofs-of-the-formulas-for-geometric-progressions.lesson>The proofs of the formulas for geometric progressions</A> 

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/Sequences-and-series/Problems-on-geometric-progressions.lesson>Problems on geometric progressions</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/Sequences-and-series/Word-problems-on-geometric-progressions.lesson>Word problems on geometric progressions</A>

in this site.