Test In Algebra Level 2

From the statement as per the question:

$\displaystyle \sum_{x=1}^{n}{x} = \sum_{x=1}^{n+5}{x}-240$

$\Longrightarrow \displaystyle \sum_{x=1}^{n+5}{x}-\sum_{x=1}^{n}{x}=240$

$\Longrightarrow \displaystyle \sum_{x=n+1}^{n+5}{x}=240$

$\Longrightarrow \dfrac{5}{2}\left\{\left(n+1\right)+\left(n+5\right)\right\}=240$

$\Longrightarrow n=45$

Let $45=\displaystyle \sum_{x=1}^{m}{x}$ .

Hence,

$45=\dfrac{m\left(m+1\right)}{2}$

$\Longrightarrow m^{2}+m-90=0$

$\Longrightarrow m=9$ or $m=-10$

Since $m$ is a positive integer.

$\therefore m=\boxed{9}$

The sum of the first (n+5) is greater by 240 than the sum of the first "n", which means (n+5) has five terms more such as N, N+1, N+2, N+3 and N+4. When we add the terms , the sum is 240; hence, N+(N+1)+(N+2)+(N+3)+(N+4)=240. Solving for N, we get 46 (other terms are 47, 48, 49 and 50). We can now say that the first "n" is the first 45 natural numbers; therefore, n=45.

To find the number of natural terms whose sum is n or 45, we use the formula for arithmetic series. We assume "x" as the number of terms and the "xth" term, so, 45=x(1+x)/2; x^2+x-90=0; (x+10)(x-9)=0; x=-10 and x=9. Only the latter satisfies the given condition, therefore 45 is the sum of the first 9 natural numbers.

Please bear with my solutions because i am not quite familiar with the format.