Question

If p,x1,x2,x3...... and q,y1,y2,y3.... form two infinite APs with common difference a and b respectively, then the locus of a point Z(M,N) where M=(1/n)[x1 + x2 + x3 +...+ xn] and N=(1/n)[y1 + y2 + y3 +....+ yn] is
is
A) a(x-p)=b(y-q)
B) p(x-a)=b(y-q)
C) p(x-p)=b(y-q)
D) b(x-p)=a(y-q)​

Answer 1

Answer:

b(x-p)=a(y-q).....(Option (D)

Step-by-step explanation:

The first A.P. is p, $x_{1},x_{2},x_{3},$...... $x_{n}$ and it has common difference a.

So, the A.P. becomes p, p+a, p+2a, p+3a,......, p+na.

The second A.P. is q, $y_{1},y_{2},y_{3},$...... $y_{n}$ and it has common difference b.

So, the A.P. becomes q, q+b, q+2b, q+3b,......, q+nb.

Now, given that, M= $\frac{1}{n}[x_{1}+x_{2}+x_{3}+ .....+x_{n}]$

                               =$\frac{1}{n}[(p+a)+(p+2a)+(p+3a)+ ...... +(p+na)]$

                               =$\frac{n}{2n}[(p+a)+(p+na)]$  

{Here we have used the formula, the sum of n terms of an A.P. series with first term A and last term L is given by $\frac{n}{2}(A+L)$}  

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

⇒ M-p = $\frac{2p+a(n+1)}{2}-p$

⇒ M-p = $\frac{a(n+1)}{2}$

⇒ $\frac{M-p}{a} =\frac{n+1}{2}$ ......... (1)

Again, given that, N= $\frac{1}{n}[y_{1}+y_{2}+y_{3}+ .....+y_{n}]$

                               =$\frac{1}{n}[(q+b)+(q+2b)+(q+3b)+ ...... +(q+nb)]$

                               =$\frac{n}{2n}[(q+b)+(q+nb)]$

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

⇒ N-q = $\frac{2q+b(n+1)}{2}-p$

⇒ N-q = $\frac{b(n+1)}{2}$

⇒ $\frac{N-q}{b} =\frac{n+1}{2}$ ......... (2)  

So, from equation (1) and (2), we can write,  

$\frac{M-p}{a}=\frac{N-q}{b}$

Hence, if Z(M,N) is a point, then the locus of the point will be given by

$\frac{x-p}{a}=\frac{y-q}{b}$ {Converting into current coordinates}

⇒b(x-p)=a(y-q).....(Option (D).(Answer)