How do you write a rule for the nth term of the geometric sequence given the two terms $a_{2} = 36, a_{4} = 576$ $a_2=36, a_4=576$ ?

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Answer 1 · 2017-05-16T08:03:45

See explanation.

For any geometric sequence $a_{n}$ $a_n$ , and natural $n, k$ $n,k$ $(n > k)$ $(n>k)$ you can write that:

$a_{n} = a_{k} \cdot q^{n - k}$ $a_n=a_k*q^(n-k)$

In this task we can write that:

$q^{2} = 16$ $q^2=16$

This leads to 2 possible values of $q$ $q$ : $q_{1} = - 4$ $q_1=-4$ and $q_{2} = 4$ $q_2=4$

Now we can calculate the first term separately for $q = - 4$ $q=-4$ and $q = 4$ $q=4$

If $q = - 4$ $q=-4$ then $a_{1} = \frac{36}{- 4} = - 9$ $a_1=36/(-4)=-9$

else if $q = 4$ $q=4$ then $a_{1} = \frac{36}{4} = 9$ $a_1=36/4=9$

So this task has 2 solutions:

$a_{n} = (- 9) \cdot {(- 4)}^{n - 1}$ $a_n=(-9)*(-4)^(n-1)$ or $a_{n} = 9 \cdot 4^{n - 1}$ $a_n=9*4^(n-1)$