In a certain increasing sequence, [m]a_1=40[/m], and for eac

Question

In a certain increasing sequence,  a_1=40 , and for each integer  n \ge 3 ,  a_n=a_{n-2}+ka_{n-1}, 
where  k  is a constant. If  a_2=56  and  a_4=122 , what is the value of  a_3 ?

A. 72
B. 80
C. 84
D. 88
E. 96

SageGMATtutor · Answer

Here's a method to solve this problem:
At the start, I would consider that problems like these usually have integer solutions, so that k is likely to be 1, 2, 3 etc. If you try out k = 1, then a3 = 96 (that's answer E), and a4 would be 56+96 = 152. Not only does this prove that E is incorrect, but is shows that k<1. That's a sign that this problem is going to be difficult.
Next, algebra looks the best way, as you can use the formula an=an−2+kan−1 and substitute n = 3 and n=4 to give two equations:
a3 = 40 + 56k
122 = 56 + ka3
However, there's another warning clue here: this is going to give a quadratic equation, either with a non-integer solution (a quadratic with k) or a large number (a quadratic with a3). It would be almost impossible to solve in under 2 minutes.
For this reason, I would pivot to testing answers at this point. Let's test B:
a3 = 80, so 80 = 40 + 56k
56k = 40
k = 40/56 = 5/7
Check this with a2 + ka3 = a4
56 + 84k = 56 + (5)(80)/7
At this point, I'd see that we're going to get a non-integer solution instead of the desired 122, so I would know that B is incorrect. However, to find if B is too big or too small, I would try to estimate, i.e. 56 + (5)(80)/7 = approx 56 + (5)(11.5) = 56 + 57 = 113, i.e. too small, so the answer must be C or D. At this point in solving I'd probably be running out of time, so I'd likely be happy with a 50:50 guess. 
If I had time, I would test C using the same approach:
a3 = 84, so 84 = 40 + 56k
56k = 44
k = 44/56 = 11/14
Check this with a2 + ka3 = a4
56 + 84k = 56 + (11)(84)/14 = 56 + (11)(6) = 56 + 66 = 122, so C is correct

This is an extremely hard problem, and would only be relevant to very top quant scorers, those aiming for 85+

kevincan · Answer

Suppose a3 = x

x = 40 + 56k
122=56 + xk

(X - 40)/56= 66/x

X2 - 40x - 56*66=0

(X -84)(x + 44)=0
x= 84

chetan2u · Answer

We have the expression  a_n=a_{n-2}+ka_{n-1},  consisting of three continous terms and k.

Thus, when we take first three terms, we have two unknowns a3 and k.
Similarly when we take three terms a2, a3 and a4, we again have two same unknowns a3 and k.
Two equations and two unknowns should give us our answer..

Now lets remove k from each equation..
 a_n=a_{n-2}+ka_{n-1}, 
 \frac{(a_n-a_{n-2})}{a_{n-1}}=k

1. First three terms
 \frac{(a_3-a_{1})}{a_{2}}=k, 
 \frac{(a_3-40)}{56}=k,

2. Three terms starting from second
 \frac{(a_4-a_{2})}{a_{3}}=k, 
 \frac{(122-56)}{a_3}=k,

From the two equations, we get
 \frac{(a_3-40)}{56} = \frac{(122-56)}{a_3}

(a_3)^2-40a_3=56×66

(a_3)^2-40a_3-56×66=0

(a_3)^2-40a_3-(7×2^3)*(2×3×11)=0

(a_3)^2-40a_3-(7×12)*(11×4)

(a_3)^2-(84-44)a_3-84*44=0

(a_3-84)(a_3+44)=0

Thus a3 is 84 or -44.

84 is the answer.

HarshavardhanR · Answer

https://gmatclub.com/forum/download/file.php?id=89431 --- Harsha GMAT-Club-Forum-5ksqorbj.png

In a certain increasing sequence, [m]a_1=40[/m], and for eac

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In a certain increasing sequence, [m]a_1=40[/m], and for eac

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

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My Rewards