Question

The common difference of the A.P.: \( a_1, a_2, \ldots, a_m \) is 13 more than the common difference of the A.P.: \( b_1, b_2, \ldots, b_n \). If \( b_{31} = -277 \), \( b_{43} = -385 \) and \( a_{78} = 327 \), then \( a_1 \) is equal to:

• A. \(16\)
• B. \(19\)
• C. \(24\)
• D. \(21\)

Hint:

When two A.P.s are related through their common differences, always find the difference first before solving for terms.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
We have two arithmetic progressions. We first find the common difference of the second AP (\(b_n\)), then use the given relation to find the common difference of the first AP (\(a_m\)), and finally find \(a_1\).
Step 2: Key Formula or Approach:
Common difference \(d = \frac{T_p - T_q}{p - q}\).
General term \(a_n = a_1 + (n-1)d\).
Step 3: Detailed Explanation:
Let \(d_b\) be the common difference of the series \(b\).
\(d_b = \frac{b_{43} - b_{31}}{43 - 31} = \frac{-385 - (-277)}{12} = \frac{-108}{12} = -9\).
The common difference of series \(a\) is \(d_a = d_b + 13 = -9 + 13 = 4\).
Given \(a_{78} = 327\):
\(a_1 + (78 - 1)d_a = 327\).
\(a_1 + 77 \times 4 = 327\).
\(a_1 + 308 = 327\).
\(a_1 = 327 - 308 = 19\).
Step 4: Final Answer:
The first term \(a_1\) is 19.