Question

The 4th, 6th and last term of GP are 10, 40,640

Answer 1

So, number of terms of the GP are 10

Answer 2

Correct Question:-

The 4th , 6th and last term of a GP are 10 , 40 , 640 respectively. If the common ratio is positive, find the first term, common ratio and number of terms in the series.

Answer:-

Given:

4th term of a GP = 10

We know that,

$\sf{nth \: term \: of \: a \: GP \:(a _{n}) = a \times {r}^{n - 1}}$

So,

$\sf{a_4=a\times{r}^{4-1}}\\\\\sf{→a_4=a{r}^{3}}\\\\\sf{→a{r}^{3}=10\:-\: equation\:(1)}$

Similarly,

$\sf{a_6=a{r}^{5}}\\\\\sf{→a{r}^{5} = 40\:-\: equation\:(2)}$

Now dividing equation (2) by (1) we get,

$→\sf{ \frac{a {r}^{5} }{a {r}^{3} } = \frac{40}{10} } \\ \\ →\sf{ \frac{ {r}^{5} }{ {r}^{3} } = 4} \\ \\ → \sf{ {r}^{2} = 4 \: \: \: \: \: \: \: \: \: \: \: \: \: ( \frac{ {a}^{m} }{ {a}^{n} } = {a}^{m - n}) } \\ \\ →\sf{r = \sqrt{4} } \\ \\ → \sf \large{r = 2}$

(given that r is positive)

Substitute r value in equation (1).

$→\sf{a {r}^{3} = 10} \\ \\ → \sf{a( {2)}^{3} = 10} \\ \\ → \sf{a = \frac{10}{8} } \\ \\→ \sf \large{a = \frac{5}{4}}$

And also given that,

nth term = 640

So,

$→\sf{a \times {r}^{n - 1} = 640} \\ \\ → \sf{ \frac{5}{4} \times {2}^{n - 1} = 640} \\ \\→ \sf{ {2}^{n - 1} = 640 \times \frac{4}{5}} \\ \\ → \sf{ {2}^{n - 1} = 512} \\ \\ →\sf{ {2}^{n - 1} = {2}^{9} } \\ \\ →\sf{ {n - 1} = 9} \\ \\ →\sf{n = 9 + 1} \\ \\ → \sf \large{n = 10}$