Question

in a gp,2nd term is 24,5th term is 1280.find common ratio.​

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Common\:ratio=3.76}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt: \implies Second \: term = 24 \\ \\ \tt: \implies Fifth \: term = 1280 \\ \\ \red{\underline \bold{To \: Find :}} \\ \tt: \implies Common \: ratio = ?$

• According to given question :

$\circ \: \tt{Let \: numbers \: of \: G.P \: be \: a,ar,a {r}^{2},a {r}^{3},..,{ar}^{n}} \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies Second \: term = 24 \\ \\ \tt: \implies ar = 24 \\ \\ \tt: \implies a = \frac{24}{r} - - - - - (1) \\ \\ \bold{For \: 5th \: term} \\ \tt: \implies a_{5} = 1280 \\ \\ \tt: \implies a {r}^{4} = 1280 \\ \\ \tt: \implies a = \frac{1280}{ {r}^{4} } - - - - - (2) \\ \\ \text{from \: (1) \: and \: (2)} \\ \tt: \implies \frac{24}{r} = \frac{1280}{ {r}^{4} } \\ \\ \tt: \implies \frac{ {r}^{4} }{r} = \frac{1280}{24} \\ \\ \tt: \implies {r}^{3} = 53.34 \\ \\ \tt: \implies r = \sqrt[3]{53.34} \\ \\ \green{\tt: \implies r =3.76}$

Answer 2

$</p><p>\tt{\huge{\pink{Hello!!! }}}$

$</p><p>\tt{\red{Given \: - }}$

$\tt{ \orange{ar = 24}}........(1)$

$\tt{ \blue{a {r}^{4} = 1280 \: }} \: ..........(2)$

Dividing (2) by (1) we Get =>

$\tt{ \purple{ \implies{ {r}^{3} = \dfrac{1280}{24}}}} \\ \\ = > \tt{ \red{r = 3.76}}$