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How do you find r and a1 for the geometric sequence: a3 = 5, a8 = 1/625?

1 Answer

Jan 25, 2016

$r = \frac{1}{5}$ $r=1/5$
$a_{1} = 125$ $a_1=125$

Explanation:

For a geometric series:
$XXX a_{m} = a_{n} \cdot r^{n - m}$ $color(white)("XXX")a_m=a_n*r^(n-m)$

We will use this general formula in two forms
$XXX a_{8} = a_{3} \cdot r^{5}$ $color(white)("XXX")a_8=a_3*r^5$
and
$XXX a_{1} = a_{3} \cdot r^{- 2}$ $color(white)("XXX")a_1=a_3*r^(-2)$

We are told $a_{8} = \frac{1}{625}$ $a_8 = 1/625$ and $a_{3} = 5$ $a_3=5$
Therefore
$XXX \frac{1}{625} = 5 \cdot r^{5}$ $color(white)("XXX")1/625 = 5*r^5$

$XXX r^{5} = \frac{1}{5 \cdot 625} = \frac{1}{5 \cdot 5 \cdot 4} = \frac{1}{5^{5}}$ $color(white)("XXX")r^5=1/(5*625) = 1/(5*5*4) = 1/(5^5)$

$XXX r = \frac{1}{5}$ $color(white)("XXX")r=1/5$

and since $a_{1} = a_{3} \cdot r^{- 2}$ $a_1=a_3*r^(-2)$
$XXX a_{1} = 5 \cdot {(\frac{1}{5})}^{- 2} = 5 \cdot 5^{2} = 125$ $color(white)("XXX")a_1= 5*(1/5)^(-2) = 5*5^2=125$

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