What is the 7th term of the geometric sequence where a1 = 1,024 and a4 = -16?

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Answer 1 · 2016-05-14T07:46:02

$a_7 = 1024 (1/-4)^(6) = 1/4$

Each term in a sequence - whether geometric or arithmetic can be given as a formula or as a value.
In a GP, $a_n = ar^(n-1)$

In this example, we have $a_1 = -16 and a_4 = 1024$ , which can be written as $ar^0 and ar^3$

Let's divide them - formulae and values:

$(ar^3)/(ar) = (-16)/(1024)$

$(cancelar^3)/(cancelar) = (-16)/1024 " " rArr r^3 = 1/(-64)$

Finding the cube root gives $r = 1/-4$ .

Now that we have $a$ and $r$ we can find the value of any term in the sequence.

$a_7 = 1024 (1/-4)^(6) = 1/4$