Question

Find the 10th term of a G.P whose 8th term is 768 and the common ratio is 2.

Answer 1

Given:

r =2 ;

8th term = 768;

$t n = a(r) {}^{n - 1}$

$a {}^{r} = 768 \: \: \: \: eq(1)$

10 th term of gp = a(r)^n-1

=a(r)^8

=(ar^7) X (r^2)

=768 X 4

=3072

therefore,10 th term of a gp is 3072..

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