If `a_1,a_2,a_3,....a_n` and `b_1,b_2,b_3,....b_n` are two arithematic progression with common difference of 2nd is two more than that of first and `b_(100)=a_(70),a_(100)=-399,a_(40)=-159` then the value of `b_1` is

Options

A. -51
B. -61
C. -81
D. 81

Answer

To solve the problem, we need to find the value of \( b_1 \) given the conditions of two arithmetic progressions \( a_n \) and \( b_n \). Let's break down the solution step by step.

### Step 1: Define the terms of the arithmetic progressions
Let:
- The first term of the first arithmetic progression \( a_1 \) be \( a_1 \).
- The common difference of the first arithmetic progression be \( d \).
- The first term of the second arithmetic progression \( b_1 \) be \( b_1 \).
- The common difference of the second arithmetic progression be \( d + 2 \) (since it is two more than that of the first).

### Step 2: Write the general terms of the sequences
The \( n \)-th term of an arithmetic progression can be expressed as:
- For the first AP:
\[ a_n = a_1 + (n-1)d \]
- For the second AP:
\[ b_n = b_1 + (n-1)(d + 2) \]

### Step 3: Use the given conditions
We have the following conditions:
1. \( b_{100} = a_{70} \)
2. \( a_{100} = -399 \)
3. \( a_{40} = -159 \)

### Step 4: Set up equations based on the conditions
From the first condition \( b_{100} = a_{70} \):
\[ b_1 + 99(d + 2) = a_1 + 69d \]
This simplifies to:
\[ b_1 + 99d + 198 = a_1 + 69d \]
Rearranging gives us:
\[ b_1 + 30d + 198 = a_1 \quad \text{(Equation 1)} \]

From the second condition \( a_{100} = -399 \):
\[ a_1 + 99d = -399 \quad \text{(Equation 2)} \]

From the third condition \( a_{40} = -159 \):
\[ a_1 + 39d = -159 \quad \text{(Equation 3)} \]

### Step 5: Solve for \( d \) and \( a_1 \)
Now, subtract Equation 3 from Equation 2:
\[ (a_1 + 99d) - (a_1 + 39d) = -399 + 159 \]
This simplifies to:
\[ 60d = -240 \]
Thus, we find:
\[ d = -4 \]

Now substitute \( d \) back into Equation 2 to find \( a_1 \):
\[ a_1 + 99(-4) = -399 \]
\[ a_1 - 396 = -399 \]
\[ a_1 = -3 \]

### Step 6: Substitute \( a_1 \) and \( d \) back into Equation 1 to find \( b_1 \)
Now substitute \( a_1 \) and \( d \) into Equation 1:
\[ b_1 + 30(-4) + 198 = -3 \]
\[ b_1 - 120 + 198 = -3 \]
\[ b_1 + 78 = -3 \]
\[ b_1 = -3 - 78 = -81 \]

### Final Answer
Thus, the value of \( b_1 \) is:
\[ \boxed{-81} \]