College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 8 - Mid-Chapter Check Point - Page 743: 18

Answer

Distance covered in $15^{th}$ seconds = 464 Total distance covered after 15 seconds = 3600

Work Step by Step

Skydiver falls distance in $1^{st}$ second = 16 Skydiver falls distance in $2^{nd}$ second = 48 Skydiver falls distance in $3^{rd}$ second = 80 Skydiver falls distance in $4^{th}$ second = 112 The distance covered in dives can be expressed in sequence as = 16, 48, 80, 112,......... From sequence we can see First term $a^{1}$ = 16 Common distance (d) = 32 Distance covered in $15^{th}$ seconds found by formula of $n^{th}$ term= $a^{1}$ + (n - 1)d = 16 + (15 - 1)32 = 16 + 448 = 464 Total distance covered after 15 seconds is found by sum formula $S_{n}$ = $\frac{n}{2}(a_{1} + a_{n})$ $S_{15}$ = $\frac{15}{2}(16 + 464)$ = $\frac{15}{2}\times 480$ = $15\times 240$ = 3600
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