What is $s_{n}$ $s_n$ of the geometric series with $a_{1} = 4$ $a_1=4$ , $a_{n} = 256$ $a_n=256$ , and $n = 4$ $n=4$ ?

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What is $s_{n}$ $s_n$ of the geometric series with $a_{1} = 4$ $a_1=4$ , $a_{n} = 256$ $a_n=256$ , and $n = 4$ $n=4$ ?

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Answer 1 · 2014-07-30T15:41:07

The sum is $340$ $340$ .

Since the $n t h$ $nth$ term is given as $256$ $256$ but $n$ $n$ is given as $4$ $4$ , that means $256$ $256$ is the $4 t h$ $4th$ term. But the $4 t h$ $4th$ term of a GP equals $a r^{3}$ $ar^3$ , where a is the first term and $r$ $r$ is the common ratio of the GP. Dividing $256$ $256$ by the first term (which is given as $4$ $4$ ) shows us that $r^{3} = \frac{256}{4} = 64$ $r^3 = 256/4 = 64$ .

If $r^{3} = 64$ $r^3 = 64$ , then the common ratio $r$ $r$ must equal $4$ $4$ as well. This gives us all the information we need to use the formula for the sum of a GP, $S = \frac{a (r^{n} - 1)}{r - 1}$ $S = (a(r^n - 1))/(r - 1)$ .

In this case, $S = \frac{4 (4^{4} - 1)}{4 - 1} = \frac{4 (256 - 1)}{3} = \frac{1020}{3} = 340$ $S = (4(4^4 - 1))/(4 - 1) = (4(256 - 1))/3 = 1020/3 = 340$ .

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What is $s_{n}$ $s_n$ of the geometric series with $a_{1} = 4$ $a_1=4$ , $a_{n} = 256$ $a_n=256$ , and $n = 4$ $n=4$ ?

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