Explanation
Given A.P is $$\dfrac{1}{3q},\dfrac{1-6q}{3q}, \dfrac{1-12q}{3q}$$
Since $$a_3=a+2d$$,
To get $${a}_{20}$$ ,
$${\textbf{Step -1: Write three digit multiple of 6 as A.P. having common difference as 6.}}$$
$${\text{Multiple of 6 between 100 and 999 are as follows:}}$$
$$102,315,324, \ldots ,996$$
$${\text{So, the formed A}}{\text{.P}}{\text{. is}}$$ $$102,315,324, \ldots ,996$$
$${\text{Where,}}$$ $$a = 102:$$ $${\text{First term of A}}{\text{.P}}{\text{.}}$$
$$d = 6:$$ $${\text{common difference of A}}{\text{.P}}{\text{.}}$$
$$l = 996:$$ $${\text{Last term of A}}{\text{.P}}{\text{.}}$$
$${\text{Now we know that, formula for }}{{\text{n}}^{{\text{th}}}}{\text{ term of an A}}{\text{.P}}{\text{. is given by,}}$$
$${a_n} = a + \left( {n - 1} \right)d \ldots \left( 1 \right)$$
$${\text{Where,}}$$ $$a = $$ $${\text{First term of A}}{\text{.P}}{\text{.}}$$
$$d = $$ $${\text{common difference of A}}{\text{.P}}{\text{.}}$$
$${a_n} = $$ $${{\text{n}}^{{\text{th}}}}{\text{term of an A}}{\text{.P}}{\text{.}}$$
$$n = $$ $${\text{Total number of terms in A}}{\text{.P}}{\text{.}}$$
$${\textbf{Step -2: Substitute the known values in equation }}\left( \mathbf1 \right)$$
$$ \Rightarrow 996 = 102 + \left( {n - 1} \right)6$$
$$ \Rightarrow 996 - 102 = \left( {n - 1} \right)6$$
$$ \Rightarrow 894 = \left( {n - 1} \right)6$$
$$ \Rightarrow \left( {n - 1} \right) = \dfrac{{894}}{6}$$
$$ \Rightarrow n - 1 = 149$$
$$ \Rightarrow n = 149 + 1$$
$$ \Rightarrow n = 150$$
$${\textbf{Therefore, option B. 150 is correct answer.}}$$
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