Saved Bookmarks
| 1. |
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. |
|
Answer» 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 Learn More: How to derive sum of n TERMS of an A.P? - Brainly.in an ap consists of 45 terms. The sum of the three middle most terms ... |
|