Correct option is (D) 6
Let \(a_1=0.004=\frac4{1000}\)
\(a_2=0.02=\frac2{100}\)
\(a_3=0.1=\frac1{10}\)
Now, \(\frac{a_2}{a_1}=\cfrac{\frac{2}{100}}{\frac{4}{1000}}\)
\(=\frac{2}{100}\times\frac{1000}{4}\)
\(=\frac{10}{2}=5\)
\(\frac{a_3}{a_2}=\cfrac{\frac{1}{10}}{\frac{2}{100}}\)
\(=\frac{1}{10}\times\frac{100}{2}\)
\(=\frac{10}{2}=5\)
Hence, 0.004, 0.02, 0.1, ......… is a geometric progression whose first term is \(a=a_1=0.004=\frac4{1000}\)
& common ratio \(=r=\frac{a_2}{a_1}=5\)
Let \(n^{th}\) term of the G.P. is 12.5.
i.e., \(a_n=12.5\)
\(\Rightarrow ar^{n-1}=12.5\)
\(\Rightarrow\frac4{1000}\times5^{n-1}=\frac{125}{10}\)
\(\Rightarrow5^{n-1}=\frac{125}{10}\times\frac{1000}{4}\)
\(=125\times25\)
\(=5^3\times5^2=5^5\)
\(\therefore n-1=5\)
\(\Rightarrow n=5+1=6\)
Hence, \(6^{th}\) term of given sequence is 6.
