The first term of geometric sequence is 512 and the common ratio is 0.5 what is the 8th term of the sequence?

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Answer 1 · 2018-05-14T18:10:18

The $8^{t h} t e r m$ $8^(th) term$ of the seqn, is : $a_{8} = 4$ $a_8=4$

We know that,

the $n^{t h} t e r m$ $n^(th) term$ of geometric sequeence is:

$a_{n} = a_{1} r^{n - 1}$ $a_n=a_1 r^(n-1)$

where $a_{1}$ $a_1$ is first term and $r$ $r$ is common ratio.

We have, $a_{1} = 512 and r = 0.5 = \frac{1}{2}$ $a_1=512 and r=0.5=1/2$

So, the $8^{t h} t e r m$ $8^(th) term$ of the seqn, is :

$a_{8} = a_{1} {(r)}^{8 - 1}$ $a_8=a_1(r)^(8-1)$

$:.a_8=512(1/2)^7=512xx1/128=4$