Question

If p, x_1,x_2,x_3.... and q,y_1,y_2,y_3.. form two infinite A.P's with common difference 'a' and 'b' respectively then the locus of P$(\alpha,\beta)$ where $\alpha = \dfrac{x_1+x_2+..+x_n}{n}, \beta = \dfrac{y_1+y_2+..y_n}{n}$ is :-

Answer 1

$\large\underline{\sf{Solution-}}$

Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

↝ nᵗʰ term of an arithmetic sequence is,

$\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{a_n\:=\:a\:+\:(n\:-\:1)\:d}}}}}} \\ \end{gathered}$

and

$\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{S_n\:=\dfrac{n}{2}( \:a\:+a_n)}}}}}} \\ \end{gathered}$

Wʜᴇʀᴇ,

• aₙ is the nᵗʰ term.

• a is the first term of the sequence.

• n is the no. of terms.

• d is the common difference.

Given that,

$\sf \:p, x_1,x_2,x_3 - - \: are \: in \: AP \: with \: common \: difference \: a$

So,

• First term of an AP = p

• Common difference of an AP = a

So,

$\rm :\longmapsto\:x_n \: is \: {(n + 1)}^{th} \: term \: of \: an \: AP$

$\rm :\implies\:x_n = p + (n + 1 - 1)a$

$\rm :\implies\:x_n = p + na - - - (1)$

Also

$\sf \:q, y_1,y_2,y_3 - - \: are \: in \: AP \: with \: common \: difference \: b$

So,

• First term of an AP = q

• Common difference of an AP = b

So,

$\rm :\longmapsto\:y_n \: is \: {(n + 1)}^{th} \: term \: of \: an \: AP$

$\rm :\implies\:y_n = q + (n + 1 - 1)b$

$\rm :\implies\:y_n = q + nb - - - (2)$

Now,

Given that,

$\rm :\longmapsto\: \alpha = \dfrac{ x_1 + x_2 + x_3 + - - - + x_n}{n}$

$\rm :\longmapsto\: \alpha = \dfrac{ \dfrac{ \cancel{n}}{2}(x_1+ x_n)}{ \cancel{n}}$

$\rm :\longmapsto\: \alpha = \dfrac{x_1 + x_n}{2}$

$\rm :\longmapsto\: \alpha = \dfrac{x_1 + p + na}{2} \: \: \: \{ \: using \: (1) \}$

$\rm :\longmapsto\: \alpha = \dfrac{a + p+ p + na}{2} \: \: \: \{ \because\:x_1 = p + a \}$

$\rm :\longmapsto\:2 \alpha = 2p + a + na$

$\rm :\longmapsto\:2 \alpha - 2p = a(1 + n)$

$\bf\implies \:n + 1 = \dfrac{2 \alpha - 2p}{a} - - - (3)$

Given that

$\rm :\longmapsto\: \beta = \dfrac{ y_1 + y_2 + y_3 + - - - + y_n}{n}$

$\rm :\longmapsto\: \beta = \dfrac{ \dfrac{ \cancel{n}}{2}(y_1+ y_n)}{ \cancel{n}}$

$\rm :\longmapsto\: \beta = \dfrac{y_1 + y_n}{2}$

$\rm :\longmapsto\: \beta = \dfrac{y_1 + q + nb}{2} \: \: \: \{ \: using \: (2) \}$

$\rm :\longmapsto\: \beta = \dfrac{b + q+ q + nb}{2} \: \: \: \{ \because\:y_1 = q + b \}$

$\rm :\longmapsto\:2 \beta = 2q + b + nb$

$\rm :\longmapsto\:2 \beta - 2q = b(1 + n)$

$\bf\implies \:n + 1 = \dfrac{2 \beta - 2q}{b} - - - (4)$

Now,

Equating equation (3) and (4), we get

$\bf\implies \:\dfrac{2 \alpha - 2p}{a}= \dfrac{2 \beta - 2q}{b}$

$\bf\implies \:\dfrac{\alpha - p}{a}= \dfrac{ \beta - q}{b}$

$\rm :\longmapsto\: b\alpha - bp = a \beta - aq$

$\rm :\longmapsto\: b\alpha - bp - a \beta + aq = 0$

$\bf\implies \:\: b\alpha - a \beta + aq - bp = 0$

$\: \: \: \: \: \: \: \: \: \underbrace{ \boxed{\bf :\implies\:Locus \: of \: ( \alpha , \beta ) \: is \: a \: line}}$

Answer 2

EXPLANATION.

If p, x₁, x₂, x₃, . . . . . . xₙ and

q, y₁, y₂, y₃, . . . . . yₙ are two infinite A.P.

Common difference = a and b.

As we know that,

General term of an A.P.

⇒ a + (a + d) + (a + 2d) + . . . .

⇒ Tₙ = a + (n - 1)d.

Using same concept in equation, we get.

⇒ p, x₁, x₂, x₃, . . . . . xₙ.

First term = a = p.

Common difference = d = b - a = a.

⇒ x₁ = p + a

⇒ x₂ = p + 2a.

⇒ x₃ = p + 3a.

⇒ x₄ = p + 4a.

. . . . .

⇒ xₙ = p + (n + 1 - 1)a.

⇒ xₙ = p + na.

⇒ q, y₁, y₂, y₃, . . . . . yₙ.

First term = a = q.

Common difference = b.

⇒ y₁ = q + b.

⇒ y₂ = q + 2b.

⇒ y₃ = q + 3b.

. . . . . .

⇒ yₙ = q + (n + 1 - 1)b.

⇒ yₙ = q + nb.

$\sf \implies \alpha = \dfrac{x_{1} + x_{2} + x_{3}+ . . . . .x_{n} } {n}$

$\sf \implies \alpha = \dfrac{1}{n} \bigg[x_{1} + x_{2} + x_{3} + . . . . . x_{n} \bigg]$

$\sf \implies \alpha = \dfrac{1}{n} \bigg[(p + a) + (p + 2a) + (p + 3a) + . . . . +(p + na) \bigg]$

$\sf \implies \alpha = \dfrac{1}{n} \bigg[ p + p + p + p . . . . n \bigg] + \bigg[ a + 2a + 3a + . . . . . a \bigg]$

$\sf \implies \alpha = \dfrac{1}{n} \bigg[np \bigg] + \bigg[a (1 + 2 + 3 + . . . . . n) \bigg]$

$\sf \implies \alpha = \dfrac{1}{n} \bigg[ np + a \bigg(\dfrac{n(n + 1)}{2} \bigg) \bigg]$

$\sf \implies \alpha = \dfrac{1}{n} \bigg[ np + a \bigg(\dfrac{n^{2}+ n }{2} \bigg) \bigg]$

$\sf \implies \alpha = \dfrac{1}{n} \times \ n \bigg[ p + a \bigg(\dfrac{n + 1}{2} \bigg) \bigg]$

$\sf \implies \alpha = p + a \bigg(\dfrac{n + 1}{2} \bigg).$

$\sf \implies \dfrac{2\alpha - 2p}{a} = n + 1 . - - - - - (1)$

$\sf \implies \beta = \dfrac{y_{1} + y_{2} + y_{3} + . . . . . y_{n}}{n}$

$\sf \implies \beta = \dfrac{1}{n} \bigg[ y_{1} + y_{2} + y_{3} + . . . . . y_{n} \bigg].$

$\sf \implies \beta = \dfrac{1}{n} \bigg[ (q + b) + (q + 2b) + (q + 3b) + .. . . . . + (q + nb) \bigg]$

$\sf \implies \beta = \dfrac{1}{n} \bigg[q + q + q + . . . .. . n \bigg] + \bigg[ b + 2b + 3b + . . . . . nb \bigg].$

$\sf \implies \beta = \dfrac{1}{n} \bigg[ nq + b(1 + 2 + 3 + . . . . . n) \bigg]$

$\sf \implies \beta = \dfrac{1}{n} \bigg[ nq + b \bigg( \dfrac{n(n + 1)}{2} \bigg) \bigg].$

$\sf \implies \beta = \dfrac{1}{n} \times n \bigg[ q + b \bigg(\dfrac{n + 1}{2} \bigg) \bigg]$

$\sf \implies \beta = q + b \bigg( \dfrac{n + 1}{2} \bigg)$

$\sf \implies \dfrac{2\beta - 2q}{b} = n + 1. - - - - - (2).$

From equation (1) & (2), we get.

$\sf \implies \dfrac{2\alpha - 2p}{a} = \dfrac{2\beta - 2q}{b}$

$\sf \implies b\alpha - bp = a\beta - aq$

$\sf \implies b\alpha - a\beta - bp + aq = 0.$