Question

which term of the GP 5,25,125,625.......is 5^10?​

Answer 1

Answer:

GP - 5, 25, 125, 625........

has first term = a = 5 and

common ratio = r= 5

Now let us assume that ;

5^10 is the nth term of given GP .

So,

$a_n = {5}^{10}$

Also we know that,

$a_n = a {r}^{n - 1} \\ \implies {5}^{10} = 5 \times {5}^{n - 1} \\ \implies {5}^{10} = {5}^{n - 1 + 1 } \\ \implies {5}^{10} = {5}^{n} \\ on \: equating \: powers\\</p><p>\: of \: both \: side \: we</p><p>\: get \\n = 10 \: \: \: \: \: \: \: \: \: \boxed{ \boxed{Answer }}$

so 5^10

will be the 10th term of given GP.

Answer 2

$\large \underline{ \underline{ \sf \: Solution : \: \: \: }}$

Given ,

first term (a) = 5

common ratio (r) = 5

Let , nth term = $\sf {(5)}^{10}$

We know that ,

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

$\to \sf {(5)}^{10} = 5 \times {(5)}^{n - 1} \\ \\ \to \sf</p><p>5 \times {(5)}^{9} =5 \times {(5)}^{n - 1} </p><p> \\ \\ \to \sf</p><p> {(5)}^{9} = {(5)}^{n - 1}$

Comparing index ( power ) of the 5 , we get

$\to \sf9 = n - 1 \\ \\ \to \sf n = 10$

Therefore , 10th term of the GP is 5^10