Question

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

Answer 1

Step-by-step explanation:

$nth \: term \: of \: gp = a {r}^{n - 1}$

$r = 2$

$a8 = 768$

$a {2}^{7} = 768$

$a = \frac{768}{128}$

$a = 6$

$a10 = a {r}^{9} = 6 \times 2 ^{9} = 3072$

so 10th term = 3072

Answer 2

Answer:

3072

Step-by-step explanation:

at first find the value of a(first term) then put the value of a and find the value of t10