Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 11 - Section 11.2 - Arithmetic Sequences - Exercise Set: 35

Answer

$S_{20} = 1220$

Work Step by Step

RECALL: (1) The sum of the first $n$ terms, $s_n$, of an arithmetic sequence can be found using the formula $S_n = \frac{n}{2}(a_1+a_n)$ where $a_1$=first term $a_n$ = $n^{th}$ term (2) The $n^{th}$ term, $a_n$, of an arithmetic sequence can be found using the formula $a_n=a_1 + d(n-1)$ where $d$=common difference $a_1$ = first term To find the sum of the first 20 terms of the sequence, we need to find the value of $a_{20}$. However, the value of $a_{20}$ can only be found if we know the value of $d$. The terms of the sequence increase by 6 so the common difference is $d=6$. The first term of the sequence is $4$ so $a-1=4$. Substitute these values into the formula in (2) above to obtain; $a_n= 4+6(n-1)$ Solve for the 20th term of the sequence to obtain: $a_{20} = 4 + 6(20-1) \\a_{20} = 4 + 6(19) \\a_{20} = 4 + 114 \\a_{20} = 118$ Solve for the sum of the first 20 terms using the formula in (1) above to obtain: $S_{20} = \frac{20}{2}(4+118) \\S_{20} = 10(122) \\S_{20} = 1220$
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