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

Answer 1

Let numbers removed by x and x+1

1+2+3+...n = n(n+1)/2

Mean = (n+1)/2

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

Mean = (n(n+1)/2 - x - x - 1 )/ n-2 = 105/4

n^2 + n - 4x - 2 = (n-2)105/2

n^2 + n - 4x - 2 - 52.5n +105 = 0

n^2 -51.5n +103-4x=0

n=[ 51.5 +- root(51.5^2 +16x - 412) ]/ 2

Now this root should give a value like xyz. 5 so that it can combine with 51.5 and make an integer

here, hit and trial comes into play.

51.5^2 - 412 = 2,240.25

2240.25 + 16x should be a perfect square

x=1 make this a square of 47.5

n = 2 and n = 99/2 (not integer)

n=2 is not possible. neither is n=99/2

so keep looking for more x

square of 48.5 - 2240.25 gives 112 which is 7*16

so put x=7

n = 1.5 and n=50

n=50 is well suited

So the possible answer is

n=50, number removed 7 and 8