Answer
80
Work Step by Step
Need to find $S = \sum_{n=1}^{10}(2J-3) = (\sum_{n=1}^{10}2J)-(\sum_{n=1}^{10}3) = 2(\sum_{n=1}^{10}J) - 30$
(Since we know from arithmetic sequence sum that $\sum_{i=1}^{n}J = \frac{n(n+1)}{2})$. Hence,
$S = 2(\sum_{n=1}^{10}J) - 30 = 2(\frac{10(10+1)}{2})) - 30 = 110 - 30 = 80$
