Show that the progression 625, 125, 25, 5, 1, $\dfrac{1}{5}$, ….. is a G.P.
Write its

(i) first term

(ii) common ratio

(iii) nth term

(iv) 10th term

Given,

625, 125, 25, 5, 1, $\dfrac{1}{5}$, ……..

$\Rightarrow \dfrac{125}{625} = \dfrac{25}{125} = \dfrac{1}{5}.$

Since, ratio between consecutive terms are equal, thus the series is in G.P.

a = 625

r = $\dfrac{125}{625} = \dfrac{1}{5}$

We know that,

nth term of a G.P. is given by,

T_n = ar^{n - 1}

T_n = $625 \times \Big(\dfrac{1}{5}\Big)^{n - 1}$

= $5^4\Big(\dfrac{1}{5^{n - 1}}\Big)$

= $5^{4 - (n - 1)}$

= 5^{5 - n}

= $\dfrac{1}{5^{n - 5}}$.

10th term,

T₁₀ = 5^{5 - 10}

= 5^-5

= $\dfrac{1}{5^5}$

= $\dfrac{1}{3125}$.

Hence, a = 625, r = $\dfrac{1}{5}$, T_n = $\dfrac{1}{5^{n - 5}}$, T₁₀ = $\dfrac{1}{3125}$.