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 largest three digit integer that is common to both these sequences, is

• A. 937
• B. 1037
• C. 967
• D. None of Above

Answer 1

The Correct Option is C

Given:

\( a_n = 13 + 6(n - 1) \) 
\( \Rightarrow a_n = 13 + 6n - 6 = 7 + 6n \)

Similarly, \( b_n = 15 + 7(n - 1) \)
\( \Rightarrow b_n = 15 + 7n - 7 = 8 + 7n \)

The common differences are 6 and 7 respectively.
LCM of 6 and 7 is \( \text{LCM}(6, 7) = 42 \)

The first common term is found by inspection to be 43.

So, we form a new AP starting at 43 with common difference 42: 
\( t_m = 43 + (m - 1) \cdot 42 \)

We need the largest term less than 1000:
\( 43 + (m - 1) \cdot 42 < 1000 \)
\( \Rightarrow (m - 1) \cdot 42 < 957 \)
\( \Rightarrow m - 1 < \frac{957}{42} \approx 22.78 \)
\( \Rightarrow m = 23 \)

Therefore, the 23rd term is:
\( t_{23} = 43 + (23 - 1) \cdot 42 = 43 + 22 \cdot 42 = 967 \)

Correct option: (C) 967