## Intermediate Algebra for College Students (7th Edition)

Published by Pearson

# Chapter 11 - Section 11.2 - Arithmetic Sequences - Exercise Set - Page 839: 38

#### Answer

$S_{50}=6,600$

#### 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 50 terms of the sequence, we need to find the value of $a_{50}$. However, the value of $a_{50}$ can only be found if we know the value of $d$. Solve for $d$ by subtracting the first term to the second term to obtain: $d=-9-(-15) \\d=-9+15 \\d=6$ The first term of the sequence is $-15$ so $a_1=-15$. Substitute these values into the formula in (2) above to obtain; $a_n= -15+6(n-1)$ Solve for the 50th term of the sequence to obtain: $a_{50} = -15 + 6(50-1) \\a_{50} = -15+ 6(49) \\a_{50} = -15 + 294 \\a_{50} = 279$ Solve for the sum of the first 50 terms using the formula in (1) above to obtain: $S_{50} = \frac{50}{2}(-15+279) \\S_{50} = 25(264) \\S_{50} = 6,600$

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