Question

Which term of the progression 0.004, 0.02, 0.1, .... is 12.5 ?

Answer 1

We have,

$\frac{a_2}{a_1}=\frac{0.02}{0.004}=5,\frac{a_3}{a_2}=\frac{0.1}{0.02}=5$

$\Rightarrow \frac{a_2}{a_1}=\frac{a_3}{a_2}=5$

The given progression is a G.P. whose first term, a is 0.004 and common ratio, r is 5.

Let the nth term be 12.5

$\therefore a_n=12.5$

$\Rightarrow a{r}^{n-1}=12.5$

$\Rightarrow \left(0.004\right)(5{)}^{n-1}=12.5$

$\Rightarrow (5{)}^{n-1}=\frac{12.5}{0.004}$

$\Rightarrow (5{)}^{n-1}=3125$

$\Rightarrow (5{)}^{n-1}=(5{)}^5$

Comparing the power of both the sides

$\Rightarrow n-1=.5$

$\Rightarrow n=6$

Thus, 6th term of the given G.P. is 12.5