Arithmetic Progressions Test - 53

First term: $$a = -11$$
Last term, $$l = 137$$
$$d = 4$$

$$S_{n}=\dfrac{n}{2}[t_1+t_n]$$
$$\Rightarrow 322.5=\dfrac{15}{2}[4+t_n]$$
$$\Rightarrow 322.5=7.5[4+t_n]$$
$$\Rightarrow 43=4+t_n$$
$$\Rightarrow t_n=39$$

In a arithmetic series $${ a }_{ n }=a+(n-1)d$$ where $$a$$ is the first term of the sequence, $$d$$ is the common difference, and $$n$$ is the number of the term to find.

Therefore, $$6 = a + (14 - 1) d = a + 13d$$    ....(1)
and $$14 = a + (6 - 1)d = a + 5d$$    ....(2)

According to problem:

$${\textbf{Step -1: Write three digit multiple of 6 as A.P. having common difference as 6.}}$$

               $${\text{Multiple of 6 between 100 and 999 are as follows:}}$$

               $$102,315,324, \ldots ,996$$

               $${\text{So, the formed A}}{\text{.P}}{\text{. is}}$$ $$102,315,324, \ldots ,996$$

               $${\text{Where,}}$$ $$a = 102:$$ $${\text{First term of A}}{\text{.P}}{\text{.}}$$

               $$d = 6:$$ $${\text{common difference of A}}{\text{.P}}{\text{.}}$$

               $$l = 996:$$ $${\text{Last term of A}}{\text{.P}}{\text{.}}$$

               $${\text{Now we know that, formula for }}{{\text{n}}^{{\text{th}}}}{\text{ term of an A}}{\text{.P}}{\text{. is given by,}}$$

               $${a_n} = a + \left( {n - 1} \right)d \ldots \left( 1 \right)$$

               $${\text{Where,}}$$ $$a = $$ $${\text{First term of A}}{\text{.P}}{\text{.}}$$ 

               $$d = $$ $${\text{common difference of A}}{\text{.P}}{\text{.}}$$  

               $${a_n} = $$ $${{\text{n}}^{{\text{th}}}}{\text{term of an A}}{\text{.P}}{\text{.}}$$ 

               $$n = $$ $${\text{Total number of terms in A}}{\text{.P}}{\text{.}}$$

$${\textbf{Step -2: Substitute the known values in equation }}\left( \mathbf1 \right)$$

               $$ \Rightarrow 996 = 102 + \left( {n - 1} \right)6$$

               $$ \Rightarrow 996 - 102 = \left( {n - 1} \right)6$$

               $$ \Rightarrow 894 = \left( {n - 1} \right)6$$

               $$ \Rightarrow \left( {n - 1} \right) = \dfrac{{894}}{6}$$

               $$ \Rightarrow n - 1 = 149$$

               $$ \Rightarrow n = 149 + 1$$

               $$ \Rightarrow n = 150$$

$${\textbf{Therefore, option B. 150 is correct answer.}}$$

Let $$a_1=x$$
Given, $$a_3=x+6a$$
Let $$n = 3$$
We know, $$a_n=a_1+(n-1)d$$
$$\therefore a_3=x+(3-1)d$$
$$\therefore x+6a-x=2d$$
$$\therefore \dfrac{6a}{2}=d$$
$$\therefore d=3a$$
Now use $$a_1=x, d = 3a, n = 10$$ in $$a_n=a_1+(n-1)d$$
$$\therefore a_{10}=x+(10-1)3a$$
$$\therefore a_{10}=x+27a$$