Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 11 - Further Topics in Algebra - 11.2 Arithmetic Sequences and Series - 11.2 Exercises - Page 1026: 77

Answer

$$6150$$

Work Step by Step

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

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.