# How towrite out the first five terms of each geometric sequence here? (1) a_1 = 2, r = 4 (2) a_1 = 48, r = -1/3

Jul 3, 2018

1) The first $5$ terms are $2 , 8 , 32 , 128 , 512.$
2) The first $5$ terms are $48 , - 16 , \frac{16}{3} , - \frac{16}{9} , \frac{16}{27.}$

#### Explanation:

1) The general form of for the $n t h$ of a geometric sequence with first term ${a}_{1}$, common ratio between all terms $r$ is given by

${a}_{n} = {a}_{1} {\left(r\right)}^{n - 1}$

Given ${a}_{1} = 2 , r = 4 ,$ we plug these values into the above formula, yielding a general form of the geometric sequence which we can use to get any term(s) we want:

${a}_{n} = 2 {\left(4\right)}^{n - 1}$

We know the first term ${a}_{1} = 2.$ That has been given.

For the second term, plug in $n = 2$ into ${a}_{n} = 2 {\left(4\right)}^{n - 1} :$

${a}_{2} = 2 {\left(4\right)}^{2 - 1}$
$= 2 {\left(4\right)}^{1} = 2 \left(4\right)$
$= 8$

Repeat this process for the third, fourth, and fifth terms:

${a}_{3} = 2 {\left(4\right)}^{3 - 1}$
$= 2 {\left(4\right)}^{2}$
$= 2 \left(16\right)$
$= 32$

${a}_{4} = 2 {\left(4\right)}^{4 - 1}$
$= 2 {\left(4\right)}^{3}$
$= 2 \left(64\right)$
$= 128$

${a}_{5} = 2 {\left(4\right)}^{5 - 1}$
$= 2 {\left(4\right)}^{4}$
$= 2 \left(256\right)$
$= 512$

Thus, the first $5$ terms are $2 , 8 , 32 , 128 , 512.$

2) Again, plug in ${a}_{1} = 48 , r = - \frac{1}{3}$ into ${a}_{n} = {a}_{1} {\left(r\right)}^{n - 1}$, which gives us

${a}_{n} = 48 {\left(- \frac{1}{3}\right)}^{n - 1}$

We know the first term ${a}_{1} = 48.$ To determine the second, third, fourth, and fifth terms, simply plug in $n = 2 , n = 3 , n = 4 , n = 5$ into the above formula:

${a}_{2} = 48 {\left(- \frac{1}{3}\right)}^{2 - 1}$
$= 48 \left(- \frac{1}{3}\right) = - 16$

${a}_{3} = 48 {\left(- \frac{1}{3}\right)}^{3 - 1}$
$= 48 {\left(- \frac{1}{3}\right)}^{2}$
$= \frac{48}{9} = \frac{16}{3}$

${a}_{4} = 48 {\left(- \frac{1}{3}\right)}^{4 - 1}$
$= 48 {\left(- \frac{1}{3}\right)}^{3}$
$= - \frac{48}{27} = - \frac{16}{9}$

${a}_{5} = 48 {\left(- \frac{1}{3}\right)}^{5 - 1}$
$= 48 {\left(- \frac{1}{3}\right)}^{4}$
$= \frac{48}{81} = \frac{16}{27}$

Thus, the first $5$ terms are $48 , - 16 , \frac{16}{3} , - \frac{16}{9} , \frac{16}{27.}$