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Question
Which term of the progression 0.004, 0.02, 0.1, ... is 12.5?
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Solution
We have,
\[\frac{a_2}{a_1} = \frac{0 . 02}{0 . 004} = 5, \frac{a_3}{a_2} = \frac{0 . 1}{0 . 02} = 5\]
\[ \Rightarrow \frac{a_2}{a_1} = \frac{a_3}{a_2} = 5\]
\[\text { The given progression is a G . P . whose first term, a is 0 . 004 and common ratio, r is 5 }. \]
\[\text { Let the nth term be } 12 . 5 . \]
\[ \therefore a_n = 12 . 5\]
\[ \Rightarrow a r^{n - 1} = 12 . 5\]
\[ \Rightarrow (0 . 004)(5 )^{n - 1} = 12 . 5\]
\[ \Rightarrow (5 )^{n - 1} = \frac{12 . 5}{0 . 004}\]
\[ \Rightarrow (5 )^{n - 1} = 3125\]
\[ \Rightarrow (5 )^{n - 1} = (5 )^5 \]
\[\text { Comparing the power of both the sides }\]
\[ \Rightarrow n - 1 = 5\]
\[ \Rightarrow n = 6\]
\[\text { Thus, 6th term of the given G . P . is } 12 . 5\]
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