Solve the sum of 8 terms in a geometric series?

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Solve the sum of 8 terms in a geometric series?

Determine the sum of the first eight terms of the geometric series in which:

t1 = 42 and t9 = 10752

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Answer 1 · 2018-01-08T15:56:20

The sum of the first 8 terms is $10710$ $10710$

Explanation:

The first step would be finding the common ratio.

If we take the geometric sequence with common ratio $2$ $2$ and terms

$1, 2, 4, 8$ $1, 2, 4, 8$

If given $t_{1} = 1$ $t_1 = 1$ and $t_{4} = 8$ $t_4 = 8$ , we can say that the common ratio is used $3$ $3$ times, thus the common ratio is equivalent to $\sqrt[3]{\frac{8}{1}} = 2$ $root(3)(8/1) = 2$ , which is clear when we write out all $4$ $4$ terms.

Applying this concept to our given problem, we see that

$r = \sqrt[8]{\frac{10752}{42}} = \sqrt[8]{256} = 2$ $r = root(8)(10752/42) = root(8)(256) = 2$

We know that the sum of a geometric series is given by

$S_{n} = \frac{a (1 - r^{n})}{1 - r}$ $S_n = (a(1 - r^n))/(1 - r)$

Applying this to our problem we get

$S_{8} = \frac{42 (1 - 2^{8})}{1 - 2}$ $S_8 = (42(1 - 2^8))/(1 - 2)$

$S_{8} = - 42 (1 - 2^{8})$ $S_8 = -42(1 - 2^8)$

$S_{8} = 10710$ $S_8 = 10710$

Hopefully this helps!

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Solve the sum of 8 terms in a geometric series?

Determine the sum of the first eight terms of the geometric series in which:

t1 = 42 and t9 = 10752

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

Solve the sum of 8 terms in a geometric series?

Determine the sum of the first eight terms of the geometric series in which: t1 = 42 and t9 = 10752

1 Answer

Explanation:

Related questions

Impact of this question

Determine the sum of the first eight terms of the geometric series in which:

t1 = 42 and t9 = 10752