Answer
26 terms
Work Step by Step
See p. 855.
For the arithmetic sequence $a_{n}=a+(n-1)d$
the nth partial sum $S_{n}=\displaystyle \sum_{k=1}^{n}[a+(k-1)d]$
is given by either of the following equivalent formulas:
1. $S_{n}=\displaystyle \frac{n}{2}[2a+(n-1)d]\qquad $2. $S_{n}=n(\displaystyle \frac{a+a_{n}}{2})$
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Given $a=12, d=8,$
$a_{n}=a+(n-1)d =12+8(n-1)=8n+4 $
Also, given that $S_{n}=2700=n(\displaystyle \frac{a+a_{n}}{2})$
we find n:
$2700=n(\displaystyle \frac{12+8n+4}{2}) $
$2700=n(\displaystyle \frac{8n+16}{2})$
$2700=n(4n+8)$
$4n^{2}+8n-2700=0\qquad /\div 4$
$n^{2}+2n-675=0$
Quadratic formula:
$n=\displaystyle \frac{-2\pm\sqrt{4+2700}}{2}=\frac{-2\pm 54}{2}$
$n=-1\pm 27$
...discard the negative solution,
n=26
There are 26 terms in the sum.