# How do you write a rule for the nth term of the geometric sequence and then find a_5 given a_3=-28, r=7/3?

Mar 16, 2017

$- \frac{1372}{9}$

#### Explanation:

Let say the sequences are,
${a}_{1} , {a}_{2} , {a}_{3.} . . {a}_{n}$, where,
${a}_{1} = {a}_{1} \cdot {r}^{0} = {a}_{1} \cdot {r}^{1 - 1}$
${a}_{2} = {a}_{1} \cdot r = {a}_{1} \cdot {r}^{2 - 1}$
${a}_{3} = {a}_{2} \cdot r = \left({a}_{1} \cdot r\right) \cdot r = {a}_{1} \cdot {r}^{2} = \left({a}_{1} \cdot r\right) \cdot r = {a}_{1} \cdot {r}^{3 - 1}$
Therefore,
${T}_{n} = {a}_{1} {r}^{n - 1}$

given that ${a}_{3} = - 28 \mathmr{and} r = \frac{7}{3}$

${a}_{5} = {a}_{1} \cdot {r}^{5 - 1} = {a}_{1} \cdot {r}^{4} = {a}_{1} \cdot {r}^{2} \cdot {r}^{2}$

${a}_{3} = {a}_{1} \cdot {r}^{2}$ therefore,
${a}_{1} \cdot {r}^{2} \cdot {r}^{2} = {a}_{3} \cdot {r}^{2} = - 28 \cdot {\left(\frac{7}{3}\right)}^{2} = - \frac{1372}{9}$

Mar 16, 2017

${a}_{n} = - \frac{36}{7} {\left(\frac{7}{3}\right)}^{n - 1} , {a}_{5} = - \frac{1372}{9}$

#### Explanation:

The standard geometric sequence is.

$a , a r , a {r}^{2} , a {r}^{3} , \ldots \ldots , a {r}^{n - 1}$

where a is the first term, r is the common ratio and the nth term is

• ar^(n-1)

Each term in the sequence is obtained by multiplying the previous term by r, the common ratio.

${a}_{3} = a {r}^{2} = - 28$

$\Rightarrow a \times {\left(\frac{7}{3}\right)}^{2} = - 28$

$\Rightarrow a \times \frac{49}{9} = - 28$

$\Rightarrow a = - 28 \times \frac{9}{49} = - \frac{36}{7}$

$\Rightarrow {a}_{n} = - \frac{36}{7} {\left(\frac{7}{3}\right)}^{n - 1} \leftarrow \textcolor{red}{\text{ nth term formula}}$

$\Rightarrow {a}_{5} = - \frac{36}{7} \times {\left(\frac{7}{3}\right)}^{4} = - \frac{1372}{9}$