Answer
$$6150$$
Work Step by Step
$$\eqalign{
& \sum\limits_{j = 10}^{50} {5j} \cr
& {\text{The first few terms are}} \cr
& = 5\left( {10} \right) + 5\left( {11} \right) + 5\left( {12} \right) + \cdots \cr
& = 50 + 55 + 60 + \cdots \cr
& {\text{Thus, }}{a_1} = 50{\text{ and }}d = 5.{\text{ If the sequence started with }}j = 1 \cr
& ,{\text{there would be 50 terms}}.{\text{ Because starts at }}j = 10,{\text{ nine of}} \cr
& {\text{those terms are missing, so there are 41 terms and }}n = 41 \cr
& {\text{Using the formula }}{S_n} = \frac{n}{2}\left[ {2{a_1} + \left( {n - 1} \right)d} \right] \cr
& \sum\limits_{j = 10}^{50} {5j} = {S_{41}} = \frac{{41}}{2}\left[ {2\left( {50} \right) + \left( {41 - 1} \right)\left( 5 \right)} \right] \cr
& \sum\limits_{j = 10}^{50} {5j} = \frac{{41}}{2}\left( {100 + 200} \right) \cr
& \sum\limits_{j = 10}^{50} {5j} = 6150 \cr} $$