The second and the fifth terms of a geometric sequence are 15 and 405, respectively. Is 32,805 a term of this sequence?

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Answer 1 · 2015-11-15T14:52:54

Yes

$A_n = A_1r^(n - 1)$

$A_2 = A_1r^(2 - 1)$

$=> A_2 = A_1r = 15$

$A_5 = A_1r^(5 - 1)$

$=> A_5 = A_1r^4 = 405$

$A_1rr^3 = 405$

$=> 15r^3 = 405$

$=> r^3 = 27$

$=> r = 3$

$A_2 = 15 = A_1r$
$=> A_1*3 = 15$
$=> A_1 = 5$

For 32805 to be a term of the sequence, it must satisfy the equation

$A_n = 5*3^(n - 1)$

where $n in NN$

$32805 = 5*3^(n - 1)$

$=> 6561 = 3^(n -1)$

$=> 3^8 = 3^(n - 1)$

$=> 8 = n - 1$

$=> n = 9$

Since $9 in NN$ , 32805 is a term of the sequence.