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The sum of a geometric series whose first three terms are $8000$ , $-12000$ , and $18000$ is $57875$ . What is the last term of the series?

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Exponential growth and decay: word problems

Full solution

Q. The sum of a geometric series whose first three terms are

8000

,

-12000

, and

18000

is

57875

. What is the last term of the series?

Identify common ratio: Identify the common ratio $r$ of the geometric series by dividing the second term by the first term. $\newline$ $-12000 / 8000 = -1.5$
Check third term: Check if the third term confirms the common ratio by dividing it by the second term. $\frac{18000}{-12000} = -1.5$
Use sum formula: Use the formula for the sum of a geometric series: $S = a_1 \times \frac{1 - r^n}{1 - r}$ , where $S$ is the sum, $a_1$ is the first term, $r$ is the common ratio, and $n$ is the number of terms. $\newline$ $57875 = 8000 \times \frac{1 - (-1.5)^n}{1 - (-1.5)}$
Simplify equation: Simplify the equation to solve for $n$ . $57875 = 8000 \times \left(1 + 1.5^n\right) / 2.5$ $57875 \times 2.5 = 8000 \times \left(1 + 1.5^n\right)$ $144687.5 = 8000 + 12000 \times 1.5^n$ $136687.5 = 12000 \times 1.5^n$ $136687.5 / 12000 = 1.5^n$ $11.390625 = 1.5^n$
Take logarithm: Take the logarithm of both sides to solve for $n$ .
$\log(11.390625) = n \times \log(1.5)$
$n = \frac{\log(11.390625)}{\log(1.5)}$
$n \approx 6.999$
Since $n$ must be a whole number, round $n$ to $7$ .
Find last term: Find the last term $a_n$ of the series using the formula $a_n = a_1 \cdot r^{(n-1)}$ .
$a_n = 8000 \cdot (-1.5)^{(7-1)}$
$a_n = 8000 \cdot (-1.5)^6$
$a_n = 8000 \cdot 11.390625$
$a_n = 91125$

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