Question

which term of the g.p 5,5√5,25----is 625 and find 11th term of it​

Answer 1

Step-by-step explanation:

first term = a = 5

2nd term = ar = 5√5

$\frac{ar}{a} = \frac{5 \sqrt{5} }{5} \\ r = \sqrt{5}$

$an = a {r}^{(n - 1)} \\ 625 = 5 \times {( \sqrt{5}) }^{n - 1} \\ \frac{625}{5} = {( \sqrt{5} )}^{n - 1} \\ 125 = { (\sqrt{5} )}^{n - 1} \\ {5}^{3} = ( { \sqrt{5} )}^{n - 1} \\ ({( { \sqrt{5}) }^{2} )}^{3} = ( { \sqrt{5}) }^{n - 1} \: \: \:since \: (5 = \sqrt{5} \times \sqrt{5} = { (\sqrt{5}) }^{2} ) \\ {( \sqrt{5}) }^{6} = {( \sqrt{5}) }^{n - 1} \\ bases \: are \: same \: powers \: to \: be \: equated \\ n - 1 = 6 \\ n = 6 + 1 \\ n = 7$

625 is the 7th term of the GP

11th term

$an= a {r}^{(n - 1)} \\ a11 = a {r}^{(11 - 1)} \\ = 5 \times {( \sqrt{5}) }^{10} \\ = 5 \times {( {5}^{ \frac{1}{2} } )}^{10} \\ = 5 \times {5}^{5} \\ = {5}^{6} \\ = 15625$

hope you get your answer