Answer
46.75
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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$\displaystyle \sum_{k=0}^{10}(3+0.25k)$.
is a partial sum of an arithmetic series for which the first term is
$a= 3+0.25\cdot 0=3$ ,
and $d=0.25$.
There are n=11 terms in the sum (for k=0,1,2,...,10)
The last term is $a_{11}=3+0.25\cdot 10=5.5$.
Using formula 2,
$S_{11}=\displaystyle \frac{11}{2}(3+5.5)=46.75$.
