Answer
$$9150$$
Work Step by Step
$$\eqalign{
& \sum\limits_{j = 20}^{80} {3j} \cr
& {\text{The first few terms are}} \cr
& = 3\left( {20} \right) + 3\left( {21} \right) + 3\left( {22} \right) + \cdots \cr
& = 60 + 63 + 66 + \cdots \cr
& {\text{Thus, }}{a_1} = 60{\text{ and }}d = 3.{\text{ If the sequence started with }}j = 1 \cr
& ,{\text{there would be 80 terms}}.{\text{ Because starts at }}j = 20,{\text{ 19 of}} \cr
& {\text{those terms are missing, so there are 61 terms and }}n = 61 \cr
& {\text{Using the formula }}{S_n} = \frac{n}{2}\left[ {2{a_1} + \left( {n - 1} \right)d} \right] \cr
& \sum\limits_{j = 20}^{80} {3j} = {S_{61}} = \frac{{61}}{2}\left[ {2\left( {60} \right) + \left( {61 - 1} \right)\left( 3 \right)} \right] \cr
& \sum\limits_{j = 20}^{80} {3j} = \frac{{61}}{2}\left( {120 + 180} \right) \cr
& \sum\limits_{j = 20}^{80} {3j} = 9150 \cr} $$
