This is a rejoinder to my blog posted on 11 june.-reproduced below-
A sequence of numbers a(1),a(2),a(3)......a(100) has the property that for every integer k between 1 and k inclusive the number a(k) is k less than the sum of the other 99 numbers.Find the 50th number-
There has been no response so far.
The solution can be arrived at as follows-
If S is the sum of the 100 numbers,a(k)=S-a(k)-k.
So 2a(k)=S-k
Substituting the values 1 to 100 for k in this equation and adding all the results we will have
2S=100S-(the sum of the 100 numbers from 1 to 100) or
98S=5050 or S= 5050/98.
So a(50)=(5050/98)-a(50)-50
or 2a(50)=(5050/98)-50=150/98
or a(50)=75/98
A sequence of numbers a(1),a(2),a(3)......a(100) has the property that for every integer k between 1 and k inclusive the number a(k) is k less than the sum of the other 99 numbers.Find the 50th number-
There has been no response so far.
The solution can be arrived at as follows-
If S is the sum of the 100 numbers,a(k)=S-a(k)-k.
So 2a(k)=S-k
Substituting the values 1 to 100 for k in this equation and adding all the results we will have
2S=100S-(the sum of the 100 numbers from 1 to 100) or
98S=5050 or S= 5050/98.
So a(50)=(5050/98)-a(50)-50
or 2a(50)=(5050/98)-50=150/98
or a(50)=75/98
