## College Algebra 7th Edition

Published by Brooks Cole

# Chapter 8, Sequences and Series - Section 8.2 - Arithmetic Sequences - 8.2 Exercises - Page 607: 57

#### Answer

$S_{101}=20,301$

#### Work Step by Step

RECALL: (1) The sum of the first $n$ terms of an arithmetic sequence is given by the formula: $S_n=\dfrac{n}{2}(a+a_n)$ where $a$ = first term $d$ = common difference $a_n$ = $n^{th}$ term (2) The $n^{th}$ term $a_n$ of an arithmetic sequence is given by the formula: $a_n = a + (n-1)d$ where $a$ = first term $d$ = common difference The given arithmetic sequence has: $a=1 \\a_n = 401 \\d=5-1=4$ The formula for the partial sum requires the values of $a$, $a_n$ and $n$. However, only $a$ and $a_n$ are known at the moment. Solve for $n$ using the formula for $a_n$ to obtain: $a_n = a + (n-1)d \\401 = 1+(n-1)4 \\401-1 = (n-1)(4) \\400=(n-1)(4) \\\dfrac{400}{4}=\dfrac{(n-1)(4)}{4} \\100=n-1 \\100+1=n-1+1 \\101=n$ Now that it is known that $n=101$, the sum of the first 101 terms can be computed using the formula above. $S_{101} = \dfrac{101}{2}(1+401) \\S_{101}=\dfrac{101}{2}(402) \\S_{101}=20,301$

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