# How do you find S_n for the geometric series a_1=343, a_n=-1, r=-1/7?

Jul 10, 2017

${S}_{n} = 300$

#### Explanation:

${a}_{1} = 343 , {a}_{n} = = - 1 , r = - \frac{1}{7}$

Given that

${a}_{n} = \text{last term}$

${a}_{1} = \text{first term}$

$r = - \frac{1}{7}$

${T}_{n} = {a}_{1} \cdot {r}^{n - 1}$

Recall that ${T}_{n} = {a}_{n}$

Hence $\to$ ${a}_{n} = {a}_{1} \cdot {r}^{n - 1}$

${a}_{n} = {a}_{1} \cdot {r}^{n - 1} \mathmr{and} 343 \cdot {\left(- \frac{1}{7}\right)}^{n - 1} = - 1 \mathmr{and} {\left(- \frac{1}{7}\right)}^{n - 1} = - \frac{1}{343}$ or

${\left(- \frac{1}{7}\right)}^{n - 1} = {\left(- \frac{1}{7}\right)}^{3} \therefore n - 1 = 3 \mathmr{and} n = 4$

${S}_{n} = {a}_{1} \cdot \frac{{r}^{n} - 1}{r - 1} = 343 \cdot \frac{{\left(- \frac{1}{7}\right)}^{4} - 1}{- \frac{1}{7} - 1}$

$= 343 \cdot \frac{\frac{1}{2401} - 1}{- \frac{8}{7}} = 343 \cdot \left(- \frac{2400}{2401}\right) \cdot \left(- \frac{7}{8}\right)$

$= \cancel{343} \cdot \frac{300}{\cancel{343}} = 300$

${S}_{n} = 300$ [Ans]