How do you write a rule for the nth term of the geometric sequence and then find $a_5$ given $a_4=1/27, r=4/3$ ?

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Answer 1 · 2017-07-30T03:29:20

$a_n = a_1r^(n-1)$
$a_5 =4/81$

The $n^(th)$ term of a geometric sequence with first term $a_1$ and common ratio $r$ is given by:

$a_n = a_1r^(n-1)$

In this example, $a_4 = 1/27 and r=4/3$

$:. 1/27 = a_1 xx (4/3)^3$

$1/27 = a_1 xx 4^3/3^3$

$1/27 = a_1 xx 64/27$

$64a_1 = 27/27$

$a_1 = 1/64$

We are asked to find $a_5$ using the formula for the $n^(th)$ term above.

$a_5 = a_1 xx r^4$

$= 1/64 xx (4/3)^4$

$= 1/64 xx 256/81$

$= 4/81$

NB: We could have found this result more simply by using:

$a_n = a_(n-1) xx r$

$:. a_5 = a_4 xx r$

$a_5 = 1/27 xx 4/3 = 4/81$

How do you write a rule for the nth term of the geometric sequence and then find a_5a_5 given a_4=1/27, r=4/3a_4=1/27, r=4/3?