Find the standard deviation of 64, 65, 67, 72, 72, 73, 77, 78, 91, 96 The mean is 75.5 What is the easiest way to do this problem? Solution S.D. = sqrt((((64-75.5)^2)+((65-75.5)^2)+((67-75.5)^2)+((72-75.5)^2)+((72- 75.5)^2)+((73-75.5)^2)+((77-75.5)^2)+((78-75.5)^2)+((91-75.5)^2)+((96-75.5)^2))/10) = 10.07224.

1. find the sum of the first 50 terms of the arithmetic progression, given that the 15th term is 34 and
the sum of the first 5 terms is 20
Solution
The nth term of an arithmetic progression is [t_n = a +(n - 1)*d] where a is the first term and d is
the common difference.
The sum of the first n terms is [S_n = (n/2)*(2a + (n-1)*d)]
If the 15th term of the series is 34, [a + 14d = 34 => a = 34 - 14d]
And as the sum of the first 5 terms is 20. (5/2)(2a + 14d) = 20
=> 2(34 - 14d) + 4d = 8
=> 68 - 28d + 4d = 8
=> 24d = 60
=> d = 5/2
a = 34 - 14*(5/2) = -1
The sum of the first 50 terms is [25*(-2 + 49*(5/2)) = 6025/2]