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CAT 2023 Slot 2 — QA Question 22

Arithmetic ProgressionEasy

Let an and bn be two sequences such that an = 13 + 6(n - 1) and bn = 15 + 7(n - 1) for all natural numbers n. Then, the largest three digit integer that is common to both these sequences is

Answer & solution

Answer: 967

Solution

Easy

The terms common to two APs themselves form an AP whose common difference is the LCM of the two. Find the first shared term, build that combined AP, then take the largest three-digit member.

an=13+6(n−1)a_n=13+6(n-1): 13,19,25,31,37,43,…13,19,25,31,37,43,\dots    bn=15+7(n−1)b_n=15+7(n-1): 15,22,29,36,43,50,…15,22,29,36,43,50,\dots

1

First common term and the combined difference. Scanning both lists, 4343 is the first match; the common terms recur every lcm⁡(6,7)\operatorname{lcm}(6,7):

first common term=43common difference=lcm⁡(6,7)=42\begin{aligned} &\text{first common term}=43\\ &\text{common difference}=\operatorname{lcm}(6,7)=42 \end{aligned}
2

General common term:

cn=43+42(n−1)=42n+1\begin{aligned} &c_n=43+42(n-1)=42n+1 \end{aligned}
3

Largest three-digit value needs cn≤999c_n\le 999:

42n+1≤999⇒ n≤99842=23.7…⇒ n=23⇒ c23=42⋅23+1=967\begin{aligned} &42n+1\le 999\\ &\Rightarrow\ n\le \frac{998}{42}=23.7\ldots\\ &\Rightarrow\ n=23\\ &\Rightarrow\ c_{23}=42\cdot 23+1=967 \end{aligned}
967\mathbf{967}
CAT 2023 Slot 2 QA Q22: 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 — Solution | TheCATExam