Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 11 - Section 11.2 - Arithmetic and Geometric Sequences - Exercise Set: 22

Answer

$a_{25}= 75$

Work Step by Step

$a_{n}$ of the arithmetic sequence is $a_{n} = a_{1} + (n-1)d$ Similarly, Second term of arithmetic sequence is $a_{2} =6$ $a_{2} = a_{1} + (2-1)d$ $a_{2} = a_{1} + d$ $ a_{1} + d =6$ Equation $(1)$ Tenth term of arithmetic sequence is $a_{10} =30$ $a_{10} = a_{1} + (10-1)d$ $a_{10} = a_{1} + 9d$ $ a_{1} + 9d =30$ Equation $(2)$ Subtract Equation $(1)$ from Equation $(2)$ $ a_{1} + 9d - (a_{1} + d )=30 - 6$ $ a_{1} + 9d - a_{1} - d =24$ $8d = 24$ $d = 3$ Substituting $d$ value in Equation $(1)$ $ a_{1} + d =6$ $ a_{1} + 3 =6$ $ a_{1}= 3 $ Using $ a_{1} $ , $d$ values and $n=25$ , $a_{25} = a_{1} + (25-1)d$ $a_{25} = 3 + (24)3$ $a_{25} = 3 +72$ $a_{25} = 75$
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