<PRE><B>Question 347947</B>
note  {{{a[n]=a[1]*r^(n-1)}}}where {{{a[1]}}} is the first term and n the common ratio.  In this series the common ratio is r = -(1/2)
.
knowing the value of the nth term: find the number of terms in the series
{{{(-1/128)= (-1/2)[1]*r^(n-1)}}}
{{{(-1/128)= (-1/2)*r^(n-1)}}}
{{{(1/64)= (-1/2)^(n-1)}}}
{{{(-1/2)^6= (-1/2)^(n-1)}}}
6 = n-1
7 =n
.
finding the sum of the terms can be found by the following
{{{S[n] = a[1](1-r^n)/(1-r)}}}
substitute:

{{{sum((2/3)(1-r^n), n=1, 7 )}}}</PRE>