Question

The common difference of the A.P $b_1,b_2,\dots \dots ,b_m$ is $2$ more than the common difference of A.P $a_1,a_2,\dots \dots ,a_n$. If $a_{40}=-159,a_{100}=-399$and $b_{100}=a_{70}$, then $b_1$ is equal to.

• A. $-127$
• B. $81$
• C. $127$
• D. $-81$

Answer 1

The correct option is D

$-81$

Explanation of correct answer :

Finding the value of $b_1$ :

Let common difference of A.P $a_1,a_2,\dots \dots ,a_n$be $D_a$ and common difference of A.P $b_1,b_2,\dots \dots ,b_m$ be $D_b$.

Given, $D_b$ $=D_a+2$.

Nth term of an A.P = $a+\left(n-1\right)d$

$$\begin{gathered} a_{40}=-159 \\ \Rightarrow a_1+39D_a=-159\dots \left(1\right) \\ {a}_{100}=-399 \\ \Rightarrow a_1+99D_a=-399\dots \left(2\right) \end{gathered}$$

Subtracting $\left(1\right)$ from $\left(2\right)$ we get,

$$\begin{gathered} -60D_a=240 \\ \Rightarrow D_a=-4 \end{gathered}$$

Thus, $D_b$ $=-4+2=-2$

Substituting the value of $D_a$ in equation $\left(1\right)$, we get

$$\begin{gathered} a_1+39D_a=-159 \\ \Rightarrow a_1=-4 \end{gathered}$$

Given, $b_{100}=a_{70}$

$$\begin{gathered} \Rightarrow b_1+99D_b=a_1+69D_a \\ \Rightarrow b_1+99\left(-2\right)=-3+69(-4) \\ \Rightarrow b_1=-81 \end{gathered}$$

Thus, $b_1$ is $-81$.

Hence, the correct answer is option (D).