mohan514 wrote:
The sum of the even numbers between 1 and n is 79*80, where n is an odd
number, then n=?
The concepts required for this question are pretty basic. You should know them.
- Sum of positive consecutive integers starting from 1 is \(\frac{x(x + 1)}{2}\) where x is the number of integers.
1+2+3+4+5+...+10 = 10*11/2- Sum of positive consecutive even integers starting from 2 is \(x*(x + 1)\)where x is the number of even integers.
2+4+6+8+10 = 5*6 = 30The explanation is simple.
2+4+6+8+10 = 2(1+2+3+4+5) By taking 2 common
Sum = 2(5*6/2) = 5*6
- Sum of positive odd integers starting from 1 is \(x^2\) where x is the number of odd integers.
1+3+5+7+9+11 = \(6^2\)I can derive it in the following way: Say x = 6.
\(1+3+5+7+9+11 = (1+1)+(3+1)+(5+1)+(7+1)+(9+1)+(11 +1) - 6\)
I add and subtract x from the right side.
The right side becomes: \(2+4+6+8+10+12 - x = x(x+1) - x = x^2\)
Coming back to this question:
Sum of x even integers is x(x+1) = 79*80.
So x = 79 i.e. there are 79 even integers.
These 79 even integers lie between 1 and n. 2 will be first such even integer, next will be 4, next will be 6 and so on till we reach the last even integer 79*2 = 158. So all even integers from 2 to 158 lie between 1 and 159. So n must be 159.
n cannot be 161/163/165... because then, more even integers will lie between 1 and n.
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