The sum of the first $3$ $3$ terms in a geometric series with common ratio $3$ $3$ is $52$ $52$ . What are the values of $t_{1}$ $t_1$ and $t_{6}$ $t_6$ ?

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The sum of the first $3$ $3$ terms in a geometric series with common ratio $3$ $3$ is $52$ $52$ . What are the values of $t_{1}$ $t_1$ and $t_{6}$ $t_6$ ?

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Answer 1 · 2016-11-15T20:07:02

$t_{1} = 4$ $t_1 = 4$
$t_{6} = 972$ $t_6 = 972$

Explanation:

You start by finding $a$ $a$ in the formula $s_{n} = \frac{a (1 - r^{n})}{1 - r}$ $s_n = (a(1 - r^n))/(1 - r)$ , since we know the sum, the common ratio ( $r$ $r$ ) and the number of terms.

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

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

$- 104 = - 26 a$ $-104 = -26a$

$a = \frac{- 104}{- 26}$ $a = (-104)/(-26)$

$a = 4$ $a = 4$

Hence, the first term is $4$ $4$ . We can find the 6th term using the formula for the nth term of a geometric sequence, $t_{n} = a \times r^{n - 1}$ $t_n = a xx r^(n - 1)$ .

$t_{6} = 4 \times 3^{6 - 1}$ $t_6 = 4 xx 3^(6 - 1)$

$t_{6} = 972$ $t_6 = 972$

Hopefully this helps!

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The sum of the first $3$ $3$ terms in a geometric series with common ratio $3$ $3$ is $52$ $52$ . What are the values of $t_{1}$ $t_1$ and $t_{6}$ $t_6$ ?

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