## Intermediate Algebra (6th Edition)

Published by Pearson

# Chapter 11 - Section 11.2 - Arithmetic and Geometric Sequences - Exercise Set - Page 646: 21

#### Answer

$a_{8}= 31$

#### Work Step by Step

$a_{n}$ of the arithmetic sequence is $a_{n} = a_{1} + (n-1)d$ Similarly, $a_{4} = a_{1} + (4-1)d$ $a_{4} = a_{1} + 3d$ $a_{1} + 3d =19$ Equation $(1)$ $a_{15} = a_{1} + (15-1)d$ $a_{15} = a_{1} +14d$ $a_{1} +14d = 52$ Equation $(2)$ Subtracting Equation $(1)$ from Equation $(2)$ $a_{1} +14d - a_{1} - 3d = 52 - 19$ $11d = 33$ $d = 3$ Substituting $d$ value in Equation $(1)$ $a_{1} + 3d =19$ $a_{1} + 3(3) =19$ $a_{1} + 9 =19$ $a_{1} =19 -9$ $a_{1} =10$ Using $a_{1}$ and $d$ values, $a_{8}= a_{1} + (8-1) d$ $a_{8}= 10+7(3)$ $a_{8}= 21+10$ $a_{8}= 31$

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