Question

lf $p,x_1,x_2,x_3\dots \dots \dots$ and $q,y_1,y_2,y_3\dots \dots$ from two infinite AP's with common difference $a$ and $b$ respectively, then the locus of $P(\alpha ,\beta )$ where $\alpha =\frac{x_1+x_2\dots \dots .+x_n}{n}$, $\beta =\frac{y_1+y_2\dots \dots .+y_n}{n}$

• A. $a(x-p)=b(y-1)$
• B. $p(x-a)=q(y-b)$
• C. $p(x-p)=b|x-q|$
• D. $b(x-p)=a(y-q)$

Answer 1

The correct option is D $b(x-p)=a(y-q)$

AP${}_1$: $p,x_1,x_2,x_3,...$

Common difference given is $a$

$n^{th}$ term of this is

$T_{n_1}=x_n=p+na$

AP${}_2$: $q,y_1,y_2,y_3,...$

Common difference given is $b$

$n^{th}$ term of this is

$T_{n_2}=y_n=q+nb$

$\alpha =\frac{x_1+x_2+\cdots +x_n}{n}$

$\Rightarrow \alpha =\frac{\frac{n(p+a+p+na)}{2}}{n}$

$\Rightarrow \alpha =p+\frac{(n+1)a}{2}$

$\Rightarrow 2(\alpha -p)=(n+1)a$ ...(1)

$\Rightarrow \beta =\frac{y_1+y_2+\cdots +y_n}{n}$

$\Rightarrow \beta =\frac{\frac{n(q+b+q+nb)}{2}}{n}$

$\Rightarrow \beta =q+\frac{(n+1)b}{2}$

$\Rightarrow 2(\beta -q)=(n+1)b$ ...(2)

Dividing (1) by (2), we get

$\frac{\alpha -p}{\beta -q}=\frac{a}{b}$

$\Rightarrow b(\alpha -p)=a(\beta -q)$

Replacing $\alpha$ by $x$ and $\beta$ by $y$, we get

$\Rightarrow b(x-p)=a(y-q)$

Hence, option D is correct.