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Re: a, b, c, and d are consecutive integers such that a < b < c [#permalink]
1
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

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Re: a, b, c, and d are consecutive integers such that a < b < c [#permalink]
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