If the average (arithmetic mean) of n consecutive odd integers is 10, what is the least of the integers?

(1) The range of the n integers is 14
(2) The greatest of the n integers is 17

in a group of consecutive integers the average is also the median
10 would be the median
17 the largest

that would have to make 3 the smallest

Let me give it a try.

Say the first term of the progression, i.e. the least of the integers is a. Hence n-th term of the progression, i.e. the largest of the integers will be [a + 2(n - 1)].

Therefore, Range = Max - Min = [a + 2(n - 1)] - a = 2(n - 1)
and, Arithmetic Mean = (Max + Min)/2 = [a + 2(n - 1) + a]/2 = [a + (n - 1)]

Now, [a + (n - 1)] = 10 => (a + n) = 11

Thus, we can determine the value of a, once we know the value of n.

Statement 1: Range = 14
Hence, 2(n - 1) = 14
=> n = 8 => We can determine the value of a.

Sufficient

Statement 2: [a + 2(n - 1)] = 17
We have two equations in two unknowns. Hence, we can determine the value of a.
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