Question

Abscissae and ordinates of $n$ given points are in $AP$ with first term $a$ and common difference $1$ and $2$ respectively. If algebraic sum of perpendiculars from these given points on a variable line which always passess through the point $\left(\begin{array}{c}\frac{13}{2},11\end{array}\right)$ is zero, then the values of $a$ and $n$ is

• A. $a=2,n=10$
• B. $a=4,n=20$
• C. $a=3,n=9$
• D. $a=0,n=1$

Answer 1

The correct option is A $a=2,n=10$

Let $n$ given points be
$(a,a),(a+1,a+2),(a+2,a+4),....$
i.e. $(a+i-1,a+2(i-1\left)\right);i=1,2,3,......,n$.
Let the variable line be
$px+qy+r=0.....\left(i\right)$
Since, the algebraic sum of perpendiculars drawn from these $n$ points on the variable line $\left(i\right)$ is always zero.
$\sum \frac{p(a+i-1)+q(a+2i-2)+r}{\sqrt{p^2+q^2}}=0$
$\sum \left[\begin{array}{c}p(a+i-1)+q(a+2i-2)+r\end{array}\right]=0$
$p\sum (a+i-1)+q\sum (a+2i-2)+rn=0$
$p\sum \frac{a+i-1}{n}+q\sum \frac{a+2i-2}{n}+r=0$
Hence, the line (i) always passes through the point
$\left(\begin{array}{c}\sum \frac{a+i-1}{n},\sum \frac{a+2i-2}{n}\end{array}\right)$.
But it is given that this passes through the point
$\left(\begin{array}{c}\frac{13}{2},11\end{array}\right)$
$\sum \frac{a+i-1}{n}=\frac{13}{2}and\sum \frac{a+2i-2}{n}=11$
Solving the equation we get, $a=2andn=10$.