College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter 8, Sequences and Series - Section 8.3 - Geometric Sequences - 8.3 Exercises - Page 615: 51



Work Step by Step

The given geometric sequence has: $a_3=28 \\a_6=224$ Note that using the third term as starting or reference point, the sixth term of the sequence can be computed by multiplying the common ratio $r$ three times to the third term. Thus, $a_6=a_3 \cdot r \cdot r \cdot r \\a_6 = a_3 \cdot r^3$ Substitute the values of $a_3$ and $a_6$ into the equation above to obtain: $a_6 = a_3 \cdot r^3 \\224 = 28 \cdot r^3 \\\dfrac{224}{28} = \dfrac{28r^3}{28} \\8 = r^3 \\2^3 = r^3$ Take the cube root of both sides to obtain: $2=r$ RECALL: The partial sum $S_n$ (sum of the first $n$ terms) of a geometric sequence is given by the formula: $S_n=a\left(\dfrac{1-r^n}{1-r}\right), r\ne 1$ where $a$ = first term $r$ = common ratio As of now, only the value of $r$ is known. We need to find the value of $a$. Note that to find the value of the third term, the common ratio $r$ has to be multiplied twice to the first term. This means that to find the value of the first term, you can divide the common ratio twice (or $r^2$) to the third term. Thus, $a = a_3 \div r \div r \\a = a_3 \div r^2$ Substitute the values of $a_3$ and $r$ to obtain: $a=28 \div (2^2) \\a = 28 \div 4 \\a=7$ Now that both $a$ and $r$ are known, the sum of the first 6 terms can be computed using the formula above to obtain: $\require{cancel} S_6 =7\left(\dfrac{1-2^6}{1-2}\right) \\S_6=7\left(\dfrac{1-64}{-1}\right) \\S_6=7\left(\dfrac{-63}{-1}\right) \\S_6=7\cdot 63 \\S_6=441$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.