# If the sum of the first three terms of a geometric sequence is 52, and the common ratio is 3, what are the first and sixth terms?

Aug 30, 2017

${t}_{1} = 4$
${t}_{6} = 972$

#### Explanation:

I'll answer the first question and leave the other ones up to other contributors.

The formula for the sum of the first $n$ terms of a geometric sequence with first term $a$ and common ratio $r$ is

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

We know the sum of the first $3$ terms is $52$ and that the common ratio is $3$, therefore:

$52 = \frac{a \left(1 - {3}^{3}\right)}{1 - 3}$

Solving, we get:

$52 \left(- 2\right) = a \left(1 - 27\right)$

$- 104 = - 26 a$

$a = 4$

Recall that the nth term of a geometric sequence is given by ${t}_{n} = a {r}^{n - 1}$. This means that ${t}_{6} = 4 {\left(3\right)}^{5} = 972$

Hopefully this helps!