This is a good question on sequences, it’s just not all that difficult though.
\(p_n\) = \(p_{n-1}\) + 3. This means that this is an increasing sequence with a common difference of 3.
\(q_n\) = \(q_{n-1}\) – 4, which means that this is a decreasing sequence with a common difference of 4.
Therefore, these sequences will cross each other at some point. This is the point we are trying to ascertain when we try to find the smallest value of k.
Since the values of the ‘q’ sequence are much larger, the probability that the value of k is 6 is very remote. Nevertheless, let’s check. Before that, let’s simplify the given inequality.
\(p_k\) > \(q_{k+2}\) which gives us,
\(p_{k-1}\) + 3 > \(q_{k+2}\)-1 – 4 that simplifies to
\(p_{k-1}\)>\(q_{k+1}\) -7.
Let’s check with k=6. \(p_5\) = \(p_1\) + 3*4 = 11 + 12 = 23 and \(q_7\) = \(q_3\) – 4*4 = 103 – 16 = 87. Clearly, this does not satisfy the inequality.
We can instead start by trying larger values for k. Let’s try k=14. If this turns out to be the answer, GREAT! Else, we will be able to eliminate options B, C and D in one shot and E will be the answer.
For k = 14, \(p_{13}\) = \(p_1\) + 3*12 = 11 + 36 = 47 and \(q_{15}\) = \(q_3\) – 4(12) = 103 – 48 = 55.
47 < 55 – 7. The inequality is not satisfied. As discussed, options B, C and D can be eliminated.
The correct answer has to be E.
When we put k=15, \(p_{14}\) = \(p_1\) + 3*13 = 11 + 39 = 50 and \(q_{16}\) = \(q_3\) – 4(13) = 103 – 52 = 51.
50 > 51-7 i.e. 50 > 44. Satisfied.
It’s very important to break down and analyse the question stem. Doing this, as you saw, put us at an advantage by telling us that the two sequences do not cross over for smaller values of k. Therefore, we could directly try larger values of k, which also helped us to eliminate the options containing the smaller values of k.
Hope this helps!
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Crackverbal Prep Team
www.crackverbal.com