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

Question

10293 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-05-14T07:46:02

$a_{7} = 1024 {(\frac{1}{- 4})}^{6} = \frac{1}{4}$ $a_7 = 1024 (1/-4)^(6) = 1/4$

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

In this example, we have $a_{1} = - 16 and a_{4} = 1024$ $a_1 = -16 and a_4 = 1024$ , which can be written as $a r^{0} and a r^{3}$ $ar^0 and ar^3$

Let's divide them - formulae and values:

$\frac{a r^{3}}{a r} = \frac{- 16}{1024}$ $(ar^3)/(ar) = (-16)/(1024)$

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

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.

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