CAT Quantitative Aptitude Questions | CAT Algebra - Progressions questions

The question is about Arithmetic progressions. A question on possible values again. We need to find out the possible terms of two AP 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 8: Sequence P is defined by p_n = p_n-1 + 3, p₁ = 11, Sequence Q is defined as q_n = q_n-1 – 4, q₃ = 103. If p_k > q_k+2, what is the smallest value k can take?

1. 6
2. 11
3. 14
4. 15

Video Explanation

Explanatory Answer

Method of solving this CAT Question from CAT Algebra - Progressions: An arithmetic progression in the form of a sequence!

Sequence P is an A.P. with a = 11, and common difference 3.
So, P_k = 11 + (k – 1)3.
Sequence Q is an A.P. with third term 103 and common difference – 4.
t₃ = a + 2d
103 = a + 2 (– 4) or a = 111
q_k+2 = 111 + (k +1) (– 4)
q_{k + 2} = 111 – 4k – 4 = 107 – 4k
p_k > q_{k + 2}
11 + (k–1)3 > 107 – 4k
8 + 3k > 107 – 4k
7k > 99
k > \\frac{99}{7}\\)
k has to be an integer, so smallest value k can take is 15.

The question is "what is the smallest value k can take?"

Hence the answer is "15"

Choice D is the correct answer.