Let's solve the problem using two methods:

Method 1: Theory: In case of consecutive integers or Arithmetic Series Mean of all the terms = Mean of first term and last term = mean of second term and second last term and so on...

=> The average (arithmetic mean) of a, b, c, and d = The average (arithmetic mean)of b and c

So, Answer will be C

Method 2: Solving it using Algebra:

Quantity A: a, b, c, d are consecutive integers
=> b = a + 1
=> c = b + 1 = a + 2
=> d = c + 1 = a + 3

=> Average of a, b, c and d = \(\frac{a + b + c + d}{4}\) = \(\frac{a + a + 1 + a + 2 + a + 3}{4}\) => a + \(\frac{3}{2}\)

Quantity B: Average of b and c

= \(\frac{a + 1 + a + 2}{2}\) = a + \(\frac{3}{2}\)

Clearly, Quantity A = Quantity B = a + \(\frac{3}{2}\)

So, Answer will be C
Hope it helps!

Watch the following video to learn How to Sequence problems

Hello from the GRE Prep Club BumpBot!

Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.