Question

If a_1, a_2 ,a_3, as well as b_1, b_2, b_3, are in G.P. with sáme commón ratío, then the poíńtś (a_1,b_1,). (a_2, b_2) and (a_3, b_3)

O lie on a stra¡ght line
O lie on an ellípse
O lie on círcle
O are vert¡ces of a tríangle




no copy and pàste from G0ogle ❌​

Answer 1

★ Concept :-

Here the concept of GP and Slope of a line has been used. We see that three terms of different terms are in Geometric Progression (G.P.) with common ratio. So firstly we can convert other terms in the form of first terms. Then we can apply them in the formula of slope of line and then find the answer.

Let's do it !!

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★ Formula Used :-

$\;\boxed{\sf{\pink{m\;=\;\dfrac{y_{2}\:-\:y_{1}}{x_{2}\:-\:x_{1}}}}}$

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★ Solution :-

Given,

» Terms of first G.P. = a₁, a₂, and a₃

» Terms of second G.P. = b₁ b₂ and b₃

• Let the common ratio of both G.P. be r

• Let the slope of the line formed by first and second terms be m₁

• Let the slope of the line formed by second and third terms be m₂

We know that, for a G.P.

» First term = g

» Second term = g × r (where r is common ratio)

» Third term = g × r²

• For values of first G.P. ::

Using the above relation, we get

>> a₃ = a₁ × r²

>> a₂ = a₁ × r

• For the values of second G.P. ::

Using the above relation, we get

>> b₃ = b₁ × r²

>> b₂ = b₁ × r

We know that formula that,

$\;\sf{\rightarrow\;\;m\;=\;\dfrac{y_{2}\:-\:y_{1}}{x_{2}\:-\:x_{1}}}$

• m is the slope of line and x and y are the co-ordinates of the line.

• For the slope of first line ::

First line is formed by (a₂, b₂) and (a₁, b₁).

• Here x₁ = a₁
• Here x₂ = a₂
• Here y₁ = b₁
• Here y₂ = b₂

By applying values in the formula, we get

$\;\sf{\Longrightarrow\;\;m_{1}\;=\;\dfrac{b_{2}\:-\:b_{1}}{a_{2}\:-\:a_{1}}}$

By applying the value of b₂, we get

$\;\sf{\Longrightarrow\;\;m_{1}\;=\;\dfrac{b_{1}r\:-\:b_{1}}{a_{2}\:-\:a_{1}}}$

$\;\sf{\Longrightarrow\;\;m_{1}\;=\;\dfrac{b_{1}(r\:-\:1)}{a_{2}\:-\:a_{1}}}$

By applying the value of a₂, we get

$\;\sf{\Longrightarrow\;\;m_{1}\;=\;\dfrac{b_{1}(r\:-\:1)}{a_{1}r\:-\:a_{1}}}$

$\;\sf{\Longrightarrow\;\;m_{1}\;=\;\dfrac{b_{1}(r\:-\:1)}{a_{1}(r\:-\:1)}}$

Now by cancelling the like terms, we get

$\;\bf{\Longrightarrow\;\;\red{m_{1}\;=\;\dfrac{b_{1}}{a_{1}}}}$

• For the slope of second line ::

First line is formed by (a₃, b₃) and (a₂, b₂).

• Here x₁ = a₂
• Here x₂ = a₃
• Here y₁ = b₂
• Here y₂ = b₃

By applying the values in the given formula, we get

$\;\sf{\Longrightarrow\;\;m_{2}\;=\;\dfrac{b_{3}\:-\:b_{2}}{a_{3}\:-\:a_{2}}}$

Now by applying the value of b₃, we get

$\;\sf{\Longrightarrow\;\;m_{2}\;=\;\dfrac{b_{1}r^{2}\:-\:b_{2}}{a_{3}\:-\:a_{2}}}$

Now by applying the value of b₂, we get

$\;\sf{\Longrightarrow\;\;m_{2}\;=\;\dfrac{b_{1}r^{2}\:-\:b_{1}r}{a_{3}\:-\:a_{2}}}$

$\;\sf{\Longrightarrow\;\;m_{2}\;=\;\dfrac{b_{1}(r^{2}\:-\:r)}{a_{3}\:-\:a_{2}}}$

Now by applying the value of a₃, we get

$\;\sf{\Longrightarrow\;\;m_{2}\;=\;\dfrac{b_{1}(r^{2}\:-\:r)}{a_{1}r^{2}\:-\:a_{2}}}$

Now by applying the value of a₂, we get

$\;\sf{\Longrightarrow\;\;m_{2}\;=\;\dfrac{b_{1}(r^{2}\:-\:r)}{a_{1}r^{2}\:-\:a_{1}r}}$

$\;\sf{\Longrightarrow\;\;m_{2}\;=\;\dfrac{b_{1}(r^{2}\:-\:r)}{a_{1}(r^{2}\:-\:r)}}$

By cancelling the like terms we get,

$\;\bf{\Longrightarrow\;\;\blue{m_{2}\;=\;\dfrac{b_{1}}{a_{1}}}}$

From the values of m₁ and m₂, we get

$\;\bf{\Longrightarrow\;\;\green{m_{1}\;=\;\;m_{2}\;=\;\dfrac{b_{1}}{a_{1}}}}$

Since the slopes of the two lines are equal. This means both lines coincide each other. Since they denote a line. This means the points are on a single line.

This means the point lie on a straight line.

Thus option a) lie on a straight line is correct.

$\;\underline{\boxed{\tt{\purple{Points\;\:lie\;\:on\;\:a\;\:line}}}}$

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★ More to know :-

• For G.P. ::

$\;\tt{\leadsto\;\;a_{n}\;=\;ar^{n\:-\:1}}$

$\;\tt{\leadsto\;\;S_{n}\;=\;\dfrac{a(1\:-\:r^{n})}{1\:-\:r}}$

• For A.P. ::

$\;\tt{\leadsto\;\;a_{n}\;=\;a\:+\:(n\:-\:1)d}$

$\;\tt{\leadsto\;\;S_{n}\;=\;\dfrac{n}{2}[a\:+\:a_{n}]}$

$\;\tt{\leadsto\;\;S_{n}\;=\;\dfrac{n}{2}[2a\:+\:(n\:-\:1)d]}$