CAT Quantitative Aptitude Questions | CAT Algebra - Progressions questions

The question is about two arithmetic progressions. A question on possible smallest value again. We need to find out the smallest possible value of n which satisfies the given condition. With some simple but very powerful ideas, one can cut down on a lot of working when it comes to progressions. For example, anchoring a progression around its middle term can be very useful. CAT Exam tests the idea of progression often in the CAT Quantitative Aptitude section and this could also be tested in DI LR section of the CAT Exam as a part of a puzzle.

Question 9: The sum of 2n terms of A.P. {1, 5, 9, 13…..} is greater than sum of n terms of A.P. = {56, 58, 60..…}. What is the smallest value n can take?

1. 9
2. 10
3. 12
4. 14

Video Explanation

Explanatory Answer

Method of solving this CAT Question from CAT Algebra - Progressions: Comparing two arithmetic progressions!

First A.P., a = 1, d = 4
S_2n = \\frac{2n}{2}\\) [2 * 1+(2n −1)4]
S_2n = n(2 + 8n – 4) = n(8n – 2) = 8n² – 2n
For the second AP, a = 56, d = 2
S_n = \\frac{n}{2}\\) [2 * 56+(n −1)2]
S_n = \\frac{n}{2}\\) [112+2n −2] = \\frac{2n}{2}\\) (110+2n)
Sum of 2n terms of AP {1, 5, 9, 13, …..} is greater than sum of n terms of A.P. = {56, 58, 60, …}
8n² – 2n > \\frac{n}{2}\\) (110+2n)
16n² – 4n > 110n + 2n2
14n² > 114n
7n > 57
n > \\frac{57}{7}\\)
The smallest value n can take = 9.

The question is "What is the smallest value n can take?"

Hence the answer is "9"

Choice A is the correct answer.