"A geometric sequence's first 8 terms add up..." solved and explained

generals336

Answered question

2020-11-09

A geometric sequence's first 8 terms add up to 6560. The usual ratio is three. Which term comes third in the sequence?

Answer & Explanation

hesgidiauE

Skilled2020-11-10Added 106 answers

The formula for the sum of nn terms of a geometric sequence is $S_{n} = \frac{a_{1} (1 - r^{n})}{(1 - r)}$ where $a_{1}$ is the first term and rr is the common ratio.
If the sum of the first $n = 8$ terms is $S_{8} = 6560$ and the common ratio is $r = 3$ , then:
$S_{8} = \frac{a_{1} (1 - 3^{8})}{1 - 3}$
$6560 = \frac{a_{1} (1 - 6561)}{- 2}$
$6560 = \frac{\frac{- 6560 a_{1}}{- 2}}{}$
$6560 = - 3280 a_{1}$
$- 2 = a_{1}$
Now that we know $a_{1}$ , we can find the third term. The nnth term of a geometric sequence is given by the formula $a_{n} = a_{1} r^{n} - 1$ .
Since $a_{1} = - 2, r = 3,$ and $n = 3$ for the third term, then:
$a_{3} = - 2 {(3)}^{3} - 1 = - 2 {(3)}^{2} = - 2 (9) = - 18$

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