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?

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Answer 1 · 2017-08-30T14:09:54

$t_1 = 4$
$t_6 = 972$

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

$s_n = (a(1- r^n))/(1- r)$

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

$52 = (a(1 - 3^3))/(1 - 3)$

Solving, we get:

$52(-2) = a(1- 27)$

$-104 = -26a$

$a = 4$

Recall that the nth term of a geometric sequence is given by $t_n = a r^(n - 1)$ . This means that $t_6 = 4(3)^(5) = 972$

Hopefully this helps!

If the sum of the first three terms of a geometric sequence is 5252, and the common ratio is 33, what are the first and sixth terms?