Question
Average of eleven consecutive numbers is 42. Find the
sum of smallest and largest number.Solution
Let the numbers be (a - 5), (a - 4), (a - 3), (a - 2), (a - 1), a, (a + 1), (a + 2), (a + 3), (a + 4) and (a + 5).
So, (a - 5 + a - 4 + a - 3 + a - 2 + a - 1 + a + a + 1 + a + 2 + a + 3 + a + 4 + a + 5) Ă· 11 = 42
Or, (11a/11) = 42
Or, 'a' = 42
Required sum = a - 5 + a + 5 = 2a = 2 Ă— 42 = 84
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