What is the sum of an 8-term geometric sequence if the first term is 10 and the last term is 781,250?

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What is the sum of an 8-term geometric sequence if the first term is 10 and the last term is 781,250?

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Answer 1 · 2015-11-17T20:48:30

$976560$ $976560$

Explanation:

The common ratio is $\sqrt[7]{\frac{781250}{10}} = \sqrt[7]{78125} = 5$ $root(7)(781250/10) = root(7)(78125) = 5$

$8 \sum n = 1 10 \cdot 5^{n - 1}$ $sum_(n=1)^8 10*5^(n-1)$

$= \frac{1}{4} (5 - 1) 8 \sum n = 1 10 \cdot 5^{n - 1}$ $=1/4(5-1)sum_(n=1)^8 10*5^(n-1)$

$= \frac{1}{4} (5 8 \sum n = 1 10 \cdot 5^{n - 1} - 8 \sum n = 1 10 \cdot 5^{n - 1})$ $=1/4(5 sum_(n=1)^8 10*5^(n-1) - sum_(n=1)^8 10*5^(n-1))$

$= \frac{1}{4} (9 \sum n = 2 10 \cdot 5^{n - 1} - 8 \sum n = 1 10 \cdot 5^{n - 1})$ $=1/4(sum_(n=2)^9 10*5^(n-1) - sum_(n=1)^8 10*5^(n-1))$

$=1/4(10*5^8 + color(red)(cancel(color(black)(sum_(n=2)^8 10*5^(n-1)))) - color(red)(cancel(color(black)(sum_(n=2)^8 10*5^(n-1)))) - 10)$

$=1/4(10*5^8-10) = 5/2(5^8-1) = 5/2*390624 = 976560$

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What is the sum of an 8-term geometric sequence if the first term is 10 and the last term is 781,250?

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