Answer
50 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 $a=5, d=2700,$
$a_{n}=a+(n-1)d =5+2(n-1). $
Also, given that $S_{n}=2700=\displaystyle \frac{n}{2}[2a+(n-1)d]$,
we find n:
$2700=\displaystyle \frac{n}{2}[2(5)+2(n-1)]\quad/\times 2$
$5400=n(2n+8)$
$2n^{2}+8n-5400=0\qquad/\div 2$
$n^{2}+4n-2700=0$
... factoring: two factors of $-2700$ that add to 4
are $-54$ and 50.
(or , use the quadratic formula)
$(n+50)(n-50)=0$
$n=50$ (discard the negative solution)
There are 50 terms in the sum.
