Sum of even numbers <=n

When n is odd,Sum=(n-1)/2*(n+1)/2=79*80

n=159

Pick E

sum of n numbers = n/2(2*a + (n-1)*d)
a = is the first term of the series

d= is the difference b/w consecutive numbers

so in our case sum = 79 * 80 = 6320

a = 2 (first even number)
so the equation becomes
=>n/2*(4 * (n-1)*2)
=>2n + n^2 -n = 6320
=>n^2 + n -6320 = 0
solve it for n, then you will get solutions 79,80 and we are give that n is odd so answer is 79

Basically 2+4+6+8+10+12+..............+n-1= 79*80


2(1+2+3+4+5+6+.......)=79*80

1+2+3+4+5+................= 79*40

Use Sum=n/2(2a + (n-1)d) , where a= 1st term , d= difference, n=number of terms

n/2(2*1 + (n-1)) = 79*40

Rearranging, n^2 + n - 79*80=0

From the above equation, without doing any calculations, one can see the odd answer 79.

Is n=79 correct?

Logically we ll check this.Sum =6320

Sum of even numbers bw 1 and 79 surely ll be in d range 2000 and 3000.6000 range surely is greater.