Question

In a Go the 3rd term is 24 and 6 th term is 192.Find the term 10 th.

Answer 1

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Ans:

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$a{3} = {ar}^{2} = 24...........(1)$
$a6 = {ar}^{2} = 192......(2)$

Dividing (2) by (1) we get

${ar}^{5} \\ {ar}^{2} = 192 \: by24$

${r}^{3} = 8 = {2}^{3}$

$r = 2$

substituting r=2 in (1) we get a=6..

$a10 = {ar}^{9} = 6 {(2)}^{9} = 3072$


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Answer 2

$3rd \: \: \: term \: = 24 \\ a {r}^{2} = 24 \: \: \: - - - - 1equation \\ \\ 6th \: \: term \: \: = 192 \\ a {r}^{5} = 192 \: \: \: - - 2equation$

On Dividing 2 equation by 1 equation, we get,

${r}^{3} = 8$
r³ = 2³

r = 2

As 3rd term = ar² = 24
=> a(2)² = 24
=> 4a = 24
=> a = 6

Hence,

$10th \: \: term \: = a {r}^{9} \\ => 6 {(2)}^{9} \\ => 3({2}^{10}) \\ => 3×1024$

=> 3072