## Intermediate Algebra (6th Edition)

Published by Pearson

# Chapter 11 - Section 11.2 - Arithmetic and Geometric Sequences - Exercise Set: 23

#### Answer

$a_{9}= 20$

#### Work Step by Step

$a_{n}$ of the arithmetic progression is $a_{n} = a_{1} + (n-1)d$ Second term of arithmetic progression is $a_{2} =-1$ $a_{2} = a_{1} + (2-1)d$ $a_{2} = a_{1} + d$ $a_{1} + d =-1$ Equation $(1)$ Fourth term of arithmetic progression is $a_{4} =5$ $a_{4} = a_{1} + (4-1)d$ $a_{4} = a_{1} + 3d$ $a_{1} +3d =5$ Equation $(2)$ Subtract Equation $(1)$ from Equation $(2)$ $a_{1} +3d -(a_{1} + d ) = 5 - (-1)$ $a_{1} +3d -a_{1} - d = 5 +1$ $2d=6$ $d=3$ Substituting $d$ value in Equation $(1)$ $a_{1} + d =-1$ $a_{1} + 3 =-1$ $a_{1} =-1-3$ $a_{1} =-4$ Using $a_{1}$ , $d$ values and $n=9$ , $a_{9} = a_{1} + (9-1)d$ $a_{9} = -4 + (8)3$ $a_{9} =-4+24$ $a_{9} = 20$

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