In a G.P. the third term is 24, and the sixth term is 192. Find the tenth term.

Question 1

Question 2

${ \boxed{ \red{ \bold{3072}}}}$

Explanation:

Given,

${ \green{ \sf{ {ar}^(2) = 24}}} \: { \to} \: { \tt{ {eq}^(n) (1)}}$

${ \green{ \sf{ {ar}^(5) = 192}}} \: { \to} \: { \tt{ {eq}^(n) (2)}}$

From Eqⁿ (2),

${ \green{ \sf{ {ar}^(2). {r}^(3) = 192}}}$

${ \green{ \sf{24. {r}^(3) = 192}}}$

${ \green{ \sf{ {r}^(3) = (192)/(24)}}}$

${ \green{ \sf{ {r}^(3) = 8}}}$

${ \green{ \sf{ {r}^(3) = {2}^(3)}}}$

${ \boxed{ \purple{ \sf{r = 2}}}}$

From Eqⁿ (1)

${ \blue{ \sf{ {ar}^(2) = 24}}}$

${ \blue{ \sf{a {(2)}^(2) = 24}}}$

${ \blue{ \sf{4a = 24}}}$

${ \blue{ \sf{a = (24)/(4)}}}$

${ \boxed{ \purple{ \sf{a = 6}}}}$

10th term is,

${ \orange{ \sf{ {ar}^(9) }}}$

${ \orange{ \sf{6 {(2)}^(9)}}}$

${ \orange{ \sf{6(512)}}}$

${ \bold{ = }}{ \boxed{ \red{ \bold{3072}}}}$

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