CAT 2023 Slot 2 QA Question & Solution
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
Solution
It is given that $a_{n}=13+6(n-1)$, which can be written as $a_n=13+6n-6\ =\ 7+6n$
Similarly, $b_{n}=15+7(n-1)$, which can be written as $b_n=15+7n-7\ =\ 8+7n$
The common differences are 6, and 7, respectively, The common difference of terms that exists in both series is l.c.m (6, 7) = 42
The first common term of the first two series is 43 (by inspection)
Hence, we need to find the mth term, which is less than 1000, and the largest three-digit integer, and exists in both series.
$t_m=a+\left(m-1\right)d\ <\ 1000$
=> $43+\left(m-1\right)42\ <\ 1000$
=> $\left(m-1\right)42\ <\ 957$
=> m-1 < 22.8 => m < 23.8 => m = 23
Hence, the 23rd term is $43+22\times\ 42\ =\ 967$