Question

Let \(a_n\) and \(b_n\) be two sequences such that \(a_n = 13 + 6(n-1)\) and \(b_n = 15 + 7(n-1)\) for all natural numbers n. Then, the sum of all the three-digit numbers that are common in both series is

• A. 11487
• B. 11987
• C. 11687
• D. 11615
• E. 11944

Hint:

The common terms of two arithmetic progressions form another arithmetic progression. Its first term is the first common term found by inspection, and its common difference is the LCM of the common differences of the original two progressions.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
We are given two arithmetic progressions (APs) and asked to find the sum of their common terms that are three-digit numbers. The common terms of two APs also form an AP.
Step 2: Detailed Explanation:
The first sequence is \(a_n = 13 + 6(n-1)\). This is an AP with first term \(a_1 = 13\) and common difference \(d_a = 6\). Sequence A: 13, 19, 25, 31, 37, 43, 49, ...
The second sequence is \(b_n = 15 + 7(n-1)\). This is an AP with first term \(b_1 = 15\) and common difference \(d_b = 7\). Sequence B: 15, 22, 29, 36, 43, 50, ...
Finding the common sequence:
The first common term can be found by inspection: it is 43.
The common difference of the sequence of common terms is the Least Common Multiple (LCM) of the individual common differences. \(d_{common} = \text{LCM}(d_a, d_b) = \text{LCM}(6, 7) = 42\).
So, the sequence of common terms is an AP with first term 43 and common difference 42. Common sequence C: 43, 85, 127, 169, ...
Finding the three-digit common terms:
We need to find the terms in sequence C that are between 100 and 999. Let the general term of the common sequence be \(c_k = 43 + (k-1)42\).
We need \(100 \leq c_k \leq 999\). \[ 100 \leq 43 + (k-1)42 \leq 999 \] \[ 57 \leq (k-1)42 \leq 956 \] \[ \frac{57}{42} \leq k-1 \leq \frac{956}{42} \] \[ 1.357... \leq k-1 \leq 22.76... \] \[ 2.357... \leq k \leq 23.76... \] Since k must be an integer, \(k\) ranges from 3 to 23.
So, the three-digit common terms are from the 3rd term to the 23rd term of the common sequence.
Number of terms = \(23 - 3 + 1 = 21\).
First three-digit term (for k=3): \(c_3 = 43 + (3-1)42 = 43 + 84 = 127\).
Last three-digit term (for k=23): \(c_{23} = 43 + (23-1)42 = 43 + 22 \times 42 = 43 + 924 = 967\).
Step 3: Final Answer:
Now, we calculate the sum of this new AP (from 127 to 967 with 21 terms).
The formula for the sum of an AP is \(S_n = \frac{n}{2}(first\_term + last\_term)\).
\[ S_{21} = \frac{21}{2}(127 + 967) \] \[ S_{21} = \frac{21}{2}(1094) \] \[ S_{21} = 21 \times 547 = 11487 \]