Consider the set of numbers a, 2a, 3a, ..., na where a and n are positive integers. (a) Show that the expression for the mean of this set is . (b) Let a = 4. Find the minimum value of n for which the sum of these numbers exceeds its mean by more than 100.
Answers
Given : the set of numbers a, 2a, 3a, ..., na where a and n are positive integers.
To Find : the mean of this set is
if a = 4 then minimum value of n
Solution:
set of numbers a, 2a, 3a, ..., na
Hence numbers = n
Sum = a + 2a + 3a + ______ + na
= a( 1 + 2 + 3 + ________ + n)
= a n(n + 1)/2
Mean = a n ( n + 1)/2 n
= a (n + 1)/2
the mean of this set is a (n + 1)/2
a n(n + 1)/2 ≥ a (n + 1)/2 + 100
=> a n(n + 1) ≥ a (n + 1) + 200
=> (n + 1) a(n - 1) > 200
=> ( n² - 1) a > 200
a = 4
( n² - 1) 4 > 200
=> n² - 1 > 50
=> n ² > 51
=> n = 8
minimum value of n = 8 if a = 4
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