The common ratio in a geometric sequence is 3/2, and the fifth term is 1. How do you find the first three terms?

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Answer 1 · 2016-05-22T06:28:39

$\frac{16}{81}$ $16/81$ , $\frac{8}{27}$ $8/27$ , $\frac{4}{9}$ $4/9$

The formula for finding the $n^{th}$ $n^"th"$ ( $U_{n}$ $U_n$ ) term in a geometric sequence is

$U_{n} = U_{1} \times r^{n-1}$ $U_n=U_1xxr^"n-1"$

where $r$ $r$ is the common ratio.

$U_{1} = ?$ $U_1 = ?$
$U_{5} = 1$ $U_5 = 1$
$r = \frac{3}{2}$ $r = 3/2$

$U_{5} = U_{1} \times {(\frac{3}{2})}^{5-1}$ $U_5=U_1xx(3/2)^"5-1"$

$1 = U_{1} \times {(\frac{3}{2})}^{4}$ $1=U_1xx(3/2)^"4"$

Rearrange

$\frac{1}{{(\frac{3}{2})}^{4}} = U_{1}$ $1/(3/2)^"4"=U_1$

$U_{1} = \frac{1}{\frac{81}{16}} = \frac{16}{81}$ $U_1=1/(81/16)=16/81$

$U_{2} = \frac{16}{81} \times \frac{3}{2} = \frac{8}{27}$ $U_2=16/81xx3/2=8/27$

$U_{3} = \frac{8}{27} \times \frac{3}{2} = \frac{4}{9}$ $U_3=8/27xx3/2=4/9$