Question 1012940
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find the value of n if the sum of n terms of the series 11 + 33 + 99 + ... is equal to 108 251.
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This progression is geometric with the first term 11 and the common ratio of 3.

Use the formula for the sum of the first n terms of a geometric progression:

{{{S[n]}}} = {{{a*((q^n - 1)/(q-1))}}}. 

(see the lesson <A HREF=http://www.algebra.com/algebra/homework/Sequences-and-series/Geometric-progressions.lesson>Geometric progressions</A> in this site).

Substitute here a = 11 and q = 3. You will get an equation

{{{11*((3^n - 1)/(3-1))}}} = {{{108251}}},   or


{{{(3^n - 1)/2}}} = {{{108251/11}}} = 9841,

{{{3^n}}} = 2*9841 + 1 = 19683 = 3^9.

Hence, n = 9.

<U>Answer</U>. n = 9.
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