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)
It can be expressed as a_n = 13 + 6n - 6 = 7 + 6n
In the same manner, b_n = 15 + 7(n - 1)
It can be expressed as b_n = 15 + 7n - 7 = 8 + 7n
The common difference between them are 6 and 7 respectively.
So, the common difference of terms that exists in both series is LCM of (6, 7) = 42
Now, the first common term of the first two series is 43 (which is taken by inspection)
Therefore, we require to find the m^th term, which is less than 100, and the largest 3-digit integer, which exists in both the series.
Now,
t_m = a + (m - 1)d < 1000
⇒ 43 + (m - 1)42 < 1000
⇒ (m - 1)42 < 957
So, m - 1 < 22.8
⇒ m < 23.8 = 23
Therefore, the 23rd term is :
43 + 22 × 22 = 967

So, the correct option is (C) : 967.