How do you write a rule for the nth term of the geometric term given the two terms $a_2=2, a_6=512$ ?

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Answer 1 · 2018-02-03T06:32:57

$a_n = a * r^(n-1) = (1/2) * 4^(n-1)$

Let a be the first term and r the common ration.

First term $a_1 = a = a r^0$

$a_2 = a * r = a r^(2-1) = 2$

$a_6 = a * r*r*r*r*r = ar^5 = a r^(6-1) = 512$

$a_6 / a_2 = (a r^5) / (a r) = r^4 = 512 / 2= 256$

$r = root4(256) =root4 (4^4) = (4^4)^(1/4) = 4$

$a_2 = a r = a * 4 = 2$

$a = 1/2$

Hence the $n^(th)$ term of the geometric sequence is

$a_n = a * r^(n-1) = (1/2) * 4^(n-1)$

How do you write a rule for the nth term of the geometric term given the two terms a_2=2, a_6=512a_2=2, a_6=512?