Question

the7th term of a GP is 8 times the 4th term.Find the GP when the 5th term is 48​

Answer 1

Answer:

In any GP,

Given that,

$a_5 = 48 \\ \\ \implies a {r}^{4} = 48$

where a and r are the first term and common ratio of the GP.

Also,

we know

$a_4 = a {r}^{3} \\ and \\ a_7 = a {r}^{6}$

ATQ,

$a_7 = 8 \times a_4 \\ \\ a {r}^{6} = 8 \times a {r}^{3} \\ \\ {r}^{3 } = 8 \\ \\ \implies \boxed{ \boxed {r = 2}}$

So,

$a {r}^{4} = a \times {2}^{4} \\ \\ = 16a$

As,

$a {r}^{4} = 48 \\ \\ 16a = 48 \\ \\ \implies \boxed{ \boxed{a = 3}}$

Now,

The GP will be,

$a,ar ,a {r}^{2} ,a {r}^{3} ,a {r}^{4} ........... \\ \\ 3,6,12,24,48..............$

Answer 2

Given ,

$\star \: \sf 5th \: term = a {r}^{4} = 48$

$\star \: \sf 7th \: term = 8 \: (4th \: term) \: i.e$

$\sf \pink{ \fbox{ \red{a {r}^{6} = 8 \: (a {r}^{3} ) \: }}}$

From this ,

$\sf \implies a {r}^{6} = 8(a {r}^{3} ) \\ \\ \sf \implies {r}^{3} = 8 \\ \\ \sf \implies {r}^{3} = {(2)}^{3} \\ \\ \sf \implies r = 2$

Put the value of r = 2 in $\sf a {r}^{4} = 48$ , we obtain

$\sf \implies a {(2)}^{4} = 48 \\ \\\sf \implies a \times 16 = 48 \\ \\\sf \implies a = \frac{ \cancel{48}}{ \cancel{16}} \\ \\\sf \implies a = 3$

We know that the GP can be written as :

${ \sf \fbox{ \: a \: , \: ar \: , \: a {r}^{2} \: , \: a {r}^{3} \:, \: ........ \: a {r}^{(n - 1)} \: }}$

Therefore , the required GP is :

$\pink{ \large \fbox{\fbox{ \: 3 \: , \: 6 \: , \: 12 \: , \: 24 \: , \: ......... \: }}}$