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

Answer: (C)

967

Answer 2

Given :
an = 13 + 6(n - 1)
It can be expressed as an = 13 + 6n - 6 = 7 + 6n
In the same manner, bn = 15 + 7(n - 1)
It can be expressed as bn = 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 mth term, which is less than 100, and the largest 3-digit integer, which exists in both the series.
Now,
tm = 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.