How do you find $a_{1}$ $a_1$ for the geometric series given $S_{n} = 688$ $S_n=688$ , $a_{n} = 16$ $a_n=16$ , r=-1/2?

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How do you find $a_{1}$ $a_1$ for the geometric series given $S_{n} = 688$ $S_n=688$ , $a_{n} = 16$ $a_n=16$ , r=-1/2?

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Answer 1 · 2016-10-16T23:18:24

$S_{n} = 688 a a a a_{n} = 16 a a a r = - \frac{1}{2}$ $S_n=688color(white)(aaa)a_n=16color(white)(aaa)r=-1/2$

Find $a_{1}$ $a_1$

First use the formula for a geometric sequence.

$a_{n} = a_{1} (r^{n - 1})$ $a_n=a_1(r^(n-1))$

$16 = a_{1} {(- \frac{1}{2})}^{n - 1}$ $16=a_1(-1/2)^(n-1)$

$a_{1} = \frac{16}{{(- \frac{1}{2})}^{n - 1}} a a a$ $a_1=color(red)(frac{16}{(-1/2)^(n-1)})color(white)(aaa)$ Equation 1

Next use the formula for the sum of a geometric series.

$S_{n} = a_{1} \frac{1 - r^{n}}{1 - r}$ $S_n=a_1frac{1-r^n}{1-r}$

$688 = a_{1} \frac{1 - {(- \frac{1}{2})}^{n}}{1 - - \frac{1}{2}}$ $688=a_1frac{1- (-1/2)^n}{1- -1/2}$

$688 \cdot \frac{3}{2} = a_{1} (1 - {(- \frac{1}{2})}^{n})$ $688*3/2=a_1(1-(-1/2)^n)$

$688 \cdot \frac{3}{2} = \frac{16}{{(- \frac{1}{2})}^{n - 1}} \cdot (1 - {(- \frac{1}{2})}^{n}) a a$ $688*3/2=color(red)(frac{16}{(-1/2)^(n-1)})*(1-(-1/2)^n)color(white)(aa)$ Substitute equation $a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a$ $color(white)(aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa)$ 1 for $a_{1}$ $a_1$

$688 \cdot \frac{3}{2} = \frac{16}{{(- \frac{1}{2})}^{n - 1}} - \frac{16 \cdot {(- \frac{1}{2})}^{n}}{{(- \frac{1}{2})}^{n - 1}}$ $688*3/2=frac{16}{(-1/2)^(n-1)}-frac{16*(-1/2)^n}{(-1/2)^(n-1)}$

$688 \cdot \frac{3}{2} = \frac{16}{{(- \frac{1}{2})}^{n - 1}} - 16 {(- \frac{1}{2})}^{n - (n - 1)}$ $688*3/2=frac{16}{(-1/2)^(n-1)}-16(-1/2)^(n-(n-1))$

$688 \cdot \frac{3}{2} = \frac{16}{{(- \frac{1}{2})}^{n - 1}} - 16 (- \frac{1}{2})$ $688*3/2=frac{16}{(-1/2)^(n-1)}-16(-1/2)$

$1032 = \frac{16}{{(- \frac{1}{2})}^{n - 10}} + 8$ $1032=frac[16}((-1/2)^(n-10)}+8$

$1024 = \frac{16}{{(- \frac{1}{2})}^{n - 1}}$ $1024=frac{16}{(-1/2)^(n-1)]$

${(- \frac{1}{2})}^{n - 1} = \frac{16}{1024}$ $(-1/2)^(n-1)=16/1024$

${(- \frac{1}{2})}^{n - 1} = \frac{1}{64}$ $(-1/2)^(n-1)=1/64$

${(- \frac{1}{2})}^{6} = \frac{1}{64}$ $(-1/2)^6=1/64$

$n - 1 = 6$ $n-1=6$

$n = 7$ $n=7$

Using $a_{n} = a_{1} (r^{n - 1})$ $a_n=a_1(r^(n-1))$

$16 = a_{1} {(- \frac{1}{2})}^{7 - 1}$ $16=a_1(-1/2)^(7-1)$

$16 = a_{1} (\frac{1}{64})$ $16=a_1(1/64)$

$a_{1} = 16 \cdot 64 = 1024$ $a_1=16*64=1024$

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How do you find $a_{1}$ $a_1$ for the geometric series given $S_{n} = 688$ $S_n=688$ , $a_{n} = 16$ $a_n=16$ , r=-1/2?

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How do you find a1a_1 for the geometric series given Sn=688S_n=688, an=16a_n=16, r=-1/2?

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How do you find $a_{1}$ $a_1$ for the geometric series given $S_{n} = 688$ $S_n=688$ , $a_{n} = 16$ $a_n=16$ , r=-1/2?