Question

Two consecutive numbers are removed from the progression 1, 2, 3, ...N.  the arithmetic mean of the remaining numbers is 26 1/4.  the value of n is



a. 60 b. 81



c. 50 d. cannot be determined

Answer 1

Answer:

Option (c) 50

Step-by-step explanation:

Let the numbers removed are x and x+1 (∵ numbers removed are consecutive)

Arithmetic mean of 1, 2, 3, ......., n

= n(n + 1)/2n = (n + 1)/2         (sum of n natural numbers)

Arithmetic mean of the remaining number

=[1+2+3+.....+n-x-(x+1)]/(n-2)

or, 26 + 1/4 = [n(n+1)/2 - x - x-1)/(n-2)

or, (n² + n -4x - 2)/2(n - 2) = 105/4

or, (n² + n -4x - 2)/(n - 2) = 105/2

or, 2(n² + n -4x - 2) = 105(n - 2)

or, 2n² - 103n - 8x + 206 = 0

The roots of the above quadratic equation are

n = [103 + √(8961+64x)]/4 and n = [103 - √(8961+64x)]/4

Since n is a positive integer, therefore the value of x should be such that 8961 + 64x is a perfect square

By trial and error we get

x = 7

Therefore n = 50, 1.5

but n = 1.5 is not possible

Therefore, n = 50