NinetynineCAT 2023 Slot 1 QA Q22 Progressions & Series
Question 22 of the Quantitative Ability (QA) section, CAT 2023 Slot 1, with CAT's official answer and a worked solution.
A lab experiment measures the number of organisms at 8 am every day. Starting with 2 organisms on the first day, the number of organisms on any day is equal to 3 more than twice the number on the previous day. If the number of organisms on the \(n^{\text {th }}\) day exceeds one million, then the lowest possible value of \(n\) is
TITA: type in the answer. There are no options and no negative marking.
Answer: 19 Progressions & Series TITA
Solution
- \(a_1 = 2\), \(a_n = 2a_{n-1}+3\). Then \(a_n + 3 = 2(a_{n-1}+3)\), so \(a_n + 3 = 5\cdot2^{n-1}\).
- Need \(5\cdot2^{n-1} - 3 > 10^6\), i.e. \(2^{n-1} > 200000.6\).
- \(2^{17} = 131072\) is too small, \(2^{18} = 262144\) works, so \(n-1 = 18\), \(n = 19\).
Shortcut. Adding 3 turns \(2a+3\) recursions into pure doubling.