# Question #4af87

Sep 18, 2016

The sum is $2 , 548$.

#### Explanation:

We have to start by finding the number of terms. This can be found by solving for $n$ in the formula ${t}_{n} = a \times {r}^{n - 1}$.

$1 , 701 = 7 \times {3}^{n - 1}$

$243 = {3}^{n - 1}$

$\ln \left(243\right) = \ln \left({3}^{n - 1}\right)$

$\ln 243 = \left(n - 1\right) \ln 3$

$n - 1 = \ln \frac{243}{\ln} 3$

$n - 1 = \ln \frac{{3}^{5}}{\ln} \left({3}^{1}\right)$

$n - 1 = \frac{5 \ln 3}{1 \ln 3}$

$n - 1 = 5$

$n = 6$

Notice that at the 2nd step we could have also just equated exponents.

We will now use the formula ${s}_{n} = \frac{a \left(1 - {r}^{n}\right)}{1 - r}$ to determine the sum of the first six terms of this series.

${s}_{6} = \frac{7 \left(1 - {3}^{6}\right)}{1 - 3}$

${s}_{6} = \frac{7 \left(1 - 729\right)}{- 2}$

${s}_{6} = 2 , 548$

Hence, the sum of the geometric series with the given information is $2 , 548$.

Hopefully this helps!