## Precalculus (6th Edition) Blitzer

$975$
Since, $a_n=a_1+(n-1)d$ Here, we have $a_{15}=1+(15-1)(-4)=-55$ and $b_{15}=3+(15-1)(5)=73$ The sum of an arithmetic sequence is given by: $S_n=\dfrac{n}{2}[a_1+a_n]$ Now, $\sum_{i=1}^{15}b_i-\sum_{i=1}^{15} a_i=\dfrac{15}{2}(3+73)-\dfrac{15}{2}(1-55)$ Thus, $\sum_{i=1}^{15}b_i-\sum_{i=1}^{15} a_i=975$