CAT 2023 Slot 2 Solved Paper

It is given that ${\mathit{a}}_{\mathit{n}}=13+6\left(\mathit{n}-1\right)$, which can be written as ${\mathit{a}}_{\mathit{n}}=13+6\mathit{n}-6=7+6\mathit{n}$
Similarly, ${\mathit{b}}_{\mathit{n}}=15+7\left(\mathit{n}-1\right)$, which can be written as ${\mathit{b}}_{\mathit{n}}=15+7\mathit{n}-7=8+7\mathit{n}$
The common differences are 6 , and 7, respectively, The common difference of terms that exists in both series is I.c.m $\left(6,7\right)=42$
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.




Hence, the 23rd term is $43+22\times 42=967$