# The first three terms of a geometric series are 5, 15, 45. If the nth term of this series is 10935, what is the sum of the first n terms?

Mar 30, 2017

16400

#### Explanation:

This is a geometric series with r = 3. First, find what term $10935$ is.

$5 \cdot \left({3}^{n - 1}\right) = 10935$

$\left({3}^{n - 1}\right) = 2187$

${3}^{n - 1} = {3}^{7}$

$n - 1 = 7$

$n = 8$

Now use the formula for the sum of the first $n$ terms of a geometric series:

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

${S}_{8} = \frac{5 \left(1 - {3}^{8}\right)}{1 - 3} = 16400$

Mar 30, 2017

The sum is $16 , 400$.

#### Explanation:

Step 1: Classify the sequence

Since ${t}_{2} = 3 {t}_{1}$ and ${t}_{3} = 3 {t}_{2}$, this sequence is geometric with $r = 3$.

Step 2: Find the number of terms

There is no formula we can use to evaluate the sum without knowing the number of terms. By the formula ${t}_{n} = a {\left(r\right)}^{n - 1}$, we have:

$10935 = 5 {\left(3\right)}^{n - 1}$

$2187 = {3}^{n - 1}$

${3}^{7} = {3}^{n - 1}$

$7 = n - 1$

$n = 8$

Step 3: Evaluate the sum

The formula for the sum of a geometric series is ${s}_{n} = \frac{a \left(1 - {r}^{n}\right)}{1 - r}$.

${s}_{8} = \frac{5 \left(1 - {3}^{8}\right)}{1 - 3}$

${s}_{8} = \frac{- 32800}{- 2}$

${s}_{8} = 16 , 400$

Practice Exercises

$1$. Find the sum:

$2 + 8 + 32 + 128 + \ldots + 524 , 288$

Solution

$1.$

$699 , 050$

Hopefully this helps!