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

Published by Pearson

# Chapter 11 - Section 11.1 - Sequences and Summation Notation - Exercise Set - Page 829: 45

#### Answer

$\sum_{i=1}^{13}(ar^{i-1})$

#### Work Step by Step

Since, we have $a+ar+ar^2+.....ar^{12}$ This form a geometric progression sequence whose first term is $a$ with common ratio of $r$. For a geometric progression sequence, we have $s_n=ar^{(n-1)}$ Last term becomes $ar^{(n-1)}=ar^{12}$ or, $n=13$ Hence, $a+ar+ar^2+.....ar^{12}$ can be summed up with the summation notation as: $\sum_{i=1}^{13}(ar^{i-1})$

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