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Given an equation to solve for n:
[tex]\sum ^n_{k\mathop=1}(-21+5k)=996[/tex]The expansion of the given sum is as follows:
[tex]\begin{gathered} -21+5(1)+(-21+5(2))+(-21+5(3))+\cdots+(-21+5n)=996 \\ -21n+(5+10+15+\cdots+5n)=996 \\ -21n+5(1+2+3+\cdots+n)=996 \\ -21n+5(\frac{n(n+1)}{2})=996 \\ -21n+\frac{5}{2}(n^2+n)=996 \\ -42n+5(n^2+n)=1992 \\ -42n+5n^2+5n=1992 \\ 5n^2-37n-1992=0 \end{gathered}[/tex]Now, factorise the above quadratic equation:
[tex]\begin{gathered} 5n^2+83n-120n-1992=0 \\ n(5n+83)-24(5n+83)=0 \\ (5n+83)(n-24)=0 \end{gathered}[/tex]Use zero product rule in the equation to get:
5n + 83 = 0 or n - 24 = 0 which implies n = -83/5 and n = 24.
Neglect the negative solution of the equation to get n = 24.
Thus, the answer is 24.