Question

Among the natural numbers $1$ to $49$. find the number for which the sum of all the preceeding number and the sum of all the succeeding number are equal.

Answer 1

Natural number from $1$ to $49$.

Let $k$ be the number where sum of preceding and successding number are equal.

So,$1+2+3+....k-1=k+(k+1).....49$

adding $1+2+3+........k-1$ both sides.

$2(1+2+3+.....k-1)=1+2+....49$

$\frac{2(k-1)(k-1+1)}{2}=\frac{49(49+1)}{2}{k}^2-k=1225k^2-k-1225=0k=\frac{1\pm \sqrt{1+4\left(1225\right)}}{2}k=\frac{1\pm \sqrt{4901}}{2}k=\frac{1\pm 70}{2}=\frac{71}{2}or\frac{69}{2}$

$k=35$(approx)

Required number $=35$