Question
If, for a positive integer $$n,$$ the quadratic equation, $$x\left( {x + 1} \right) + \left( {x + 1} \right)\left( {x + 2} \right) + ..... + \left( {x + \overline {n - 1} } \right)\left( {x + n} \right) = 10n$$ has two consecutive integral solutions, then $$n$$ is equal to:
A.
11
B.
12
C.
9
D.
10
Answer :
11
Solution :
We have,
$$\eqalign{
& \sum\limits_{r{\text{ }} = {\text{ }}1}^n {\left( {x{\text{ }} + {\text{ }}r - 1} \right)\left( {x{\text{ }} + {\text{ }}r} \right) = 10n} \cr
& \sum\limits_{r{\text{ }} = {\text{ }}1}^n {\left( {{x^2} + xr{\text{ }} + \left( {r - 1} \right)x{\text{ }} + {\text{ }}{r^2} - r} \right) = 10n} \cr
& \Rightarrow \,\,\sum\limits_{r{\text{ }} = {\text{ }}1}^n {\left( {{x^2} + \left( {2r - 1} \right)x{\text{ }} + {\text{ }}r\left( {r - 1} \right)} \right) = 10n} \cr
& \Rightarrow \,n{x^2} + \left\{ {1 + 3 + 5 + ..... + \left( {2n - 1} \right)} \right\}x{\text{ }} + \left\{ {1.2 + 2.3 + ..... + \left( {n - 1} \right)n} \right\} = 10n \cr
& \Rightarrow \,n{x^2} + {n^2}x + \frac{{\left( {n - 1} \right)n\left( {n + 1} \right)}}{3} = 10n \cr
& \Rightarrow \,{x^2} + nx + \frac{{{n^2} - 31}}{3} = 0 \cr} $$
Let $$\alpha $$ and $$\alpha + 1$$ be its two solutions ( $$\because $$ it has two consequtive integral solutions )
$$\eqalign{
& \Rightarrow \,\,\alpha + \left( {\alpha + 1} \right) = - n \cr
& \Rightarrow \,\,\alpha = \frac{{ - n - 1}}{2}\,\,\,\,\,\,\,\,......\left( 1 \right) \cr
& {\text{Also }}\alpha \left( {\alpha + 1} \right) = \frac{{{n^2} - 31}}{3}\,\,\,\,\,\,\,......\left( 2 \right) \cr} $$
Putting value of (1) in (2), we get
$$\eqalign{
& - \left( {\frac{{n + 1}}{2}} \right)\left( {\frac{{1 - n}}{2}} \right) = \frac{{{n^2} - 31}}{3} \cr
& \Rightarrow \,\,{n^2} = 121 \cr
& \Rightarrow \,\,n = 11 \cr} $$