Key Concepts and Formulas

• Arithmetic Progression (A.P.): A sequence where the difference between consecutive terms is constant. This constant is called the common difference ($d$).
• $n$-th Term of an A.P.: The $n$-th term ($a_n$) of an A.P. with first term $a_1$ and common difference $d$ is given by the formula: $a_n = a_1 + (n-1)d$

---

Step-by-Step Solution

Step 1: Determine the common difference ($d_a$) and first term ($a_1$) of the A.P. $a_n$.

• Objective: To find the parameters of the first arithmetic progression, we need to determine its first term ($a_1$) and its common difference ($d_a$).
• Given Information: We are given two terms of this A.P.: $a_{40} = -159$ and $a_{100} = -399$.
• Applying the $n$-th Term Formula: Using the formula $a_n = a_1 + (n-1)d_a$, we can set up a system of two linear equations: For $a_{40}$: $a_1 + (40-1)d_a = -159$ $a_1 + 39d_a = -159 \quad \text{(Equation 1)}$ For $a_{100}$: $a_1 + (100-1)d_a = -399$ $a_1 + 99d_a = -399 \quad \text{(Equation 2)}$
• Solving the System of Equations: To find $d_a$, we subtract Equation 1 from Equation 2: $(a_1 + 99d_a) - (a_1 + 39d_a) = -399 - (-159)$ $60d_a = -399 + 159$ $60d_a = -240$ $d_a = \frac{-240}{60}$ $d_a = -4$
• Finding $a_1$: Substitute the value of $d_a$ back into Equation 1: $a_1 + 39(-4) = -159$ $a_1 - 156 = -159$ $a_1 = -159 + 156$ $a_1 = -3$ So, for A.P. $a_n$, the first term $a_1 = -3$ and the common difference $d_a = -4$.

Step 2: Calculate the 70th term ($a_{70}$) of A.P. $a_n$.

• Objective: The problem states that $b_{100} = a_{70}$. Therefore, we need to find the value of $a_{70}$.
• Applying the $n$-th Term Formula: Using $a_1 = -3$, $d_a = -4$, and $n=70$: $a_{70} = a_1 + (70-1)d_a$ $a_{70} = -3 + (69)(-4)$ $a_{70} = -3 - 276$ $a_{70} = -279$

Step 3: Determine the common difference ($d_b$) of A.P. $b_n$.

• Objective: We need to find the common difference of the second arithmetic progression, $d_b$.
• Given Information: The problem states that "The common difference of the A.P. $b_1, b_2, \ldots, b_m$ is 2 more than the common difference of A.P. $a_1, a_2, \ldots, a_n$."
• Calculating $d_b$: $d_b = d_a + 2$ $d_b = -4 + 2$ $d_b = -2$ So, the common difference of A.P. $b_n$ is $d_b = -2$.

Step 4: Determine the first term ($b_1$) of A.P. $b_n$.

• Objective: This is the value we are asked to find.
• Given Information: We are given $b_{100} = a_{70}$. From Step 2, we know $a_{70} = -279$. Thus, $b_{100} = -279$.
• Applying the $n$-th Term Formula for A.P. $b_n$: The formula for the $n$-th term of A.P. $b_n$ is $b_n = b_1 + (n-1)d_b$. We use this for $n=100$: $b_{100} = b_1 + (100-1)d_b$ $b_{100} = b_1 + 99d_b$
• Substituting Known Values: Substitute $b_{100} = -279$ and $d_b = -2$: $-279 = b_1 + 99(-2)$ $-279 = b_1 - 198$
• Solving for $b_1$: $b_1 = -279 + 198$ $b_1 = -81$

The first term of A.P. $b_n$ is $-81$.

---

Common Mistakes & Tips

• Sign Errors: Be extremely careful when dealing with negative numbers, especially during subtraction and multiplication.
• Variable Confusion: Clearly distinguish between the common differences and first terms of the two different arithmetic progressions ($d_a$ vs. $d_b$, $a_1$ vs. $b_1$).
• Translating Conditions: Ensure that conditions like "2 more than" are translated correctly into mathematical equations ($d_b = d_a + 2$, not $d_a = d_b + 2$).

---

Summary

The problem requires us to find the first term of an arithmetic progression ($b_1$) given relationships between it and another arithmetic progression ($a_n$). We first determined the common difference and first term of A.P. $a_n$ using two given terms. Then, we calculated a specific term of A.P. $a_n$ that was linked to A.P. $b_n$. Using the relationship between their common differences, we found the common difference of A.P. $b_n$. Finally, we used the $n$-th term formula for A.P. $b_n$ along with the given term value and its calculated common difference to solve for its first term, $b_1$.

The final answer is $\boxed{\text{-81}}$.