
If \[{{x}_{1}},{{x}_{2}},{{x}_{3}}\] as well as \[{{y}_{1}},{{y}_{2,}}{{y}_{3}}\] are in G.P. with the same common ratio, then the points \[({{x}_{1}},{{y}_{1}})\] , \[({{x}_{2}},{{y}_{2}})\] and \[({{x}_{3}},{{y}_{3}})\] .
A.Lie on a straight line
B.Lie on a circle
C.Lie on an ellipse
D.Are vertices of triangle
Answer
507k+ views
Hint: GP stands for Geometric Progression. \[a,ar,a{{r}^{2}},a{{r}^{3}},.....\] are said to be in GP where first term is \[a\] and common ratio is \[r\] .The \[{{n}^{th}}\] term is given by
\[{{n}^{th}}term=a{{r}^{n-1}}\]
The sum of \[n\] terms is given by \[\dfrac{a(1-{{r}^{n}})}{(1-r)}\] , when \[r<1\] and \[\dfrac{a({{r}^{n}}-1)}{(r-1)}\] , when \[r>1\] .
Geometric progression is the series of non-zero numbers in which each term after the first is found by multiplying the previous number by a fixed non-zero number called the common ratio.
If all the terms of G.P. are multiplied or divided by the same non-zero constant, then it remains a G.P. with the same common ratio. The geometric mean of \[a\] and \[b\] are given by \[\sqrt{ab}\] . The reciprocals of the terms of a given G.P. form a G.P. If each term of a G.P. is raised to the same power the resulting sequence also forms a G.P.
Complete step-by-step answer:
Given that \[{{x}_{1}},{{x}_{2}},{{x}_{3}}\] and \[{{y}_{1}},{{y}_{2,}}{{y}_{3}}\] are in G.P. with the same common ratio.
Let us assume that \[{{x}_{1}}=a\] , \[{{x}_{2}}=ar\] , \[{{x}_{3}}=a{{r}^{2}}\]
And \[{{y}_{1}}=b\] , \[{{y}_{2}}=br\] , \[{{y}_{3}}=b{{r}^{2}}\]
So, \[A({{x}_{1}},{{y}_{1}})\] , \[B({{x}_{2}},{{y}_{2}})\] and \[C({{x}_{3}},{{y}_{3}})\] can be written as \[A(a,b)\] , \[B(ar,br)\] and \[C(a{{r}^{2}},b{{r}^{2}})\]
Now we need to calculate the slope
The slope of \[AB\] is given as
\[\Rightarrow \dfrac{(b-br)}{(a-ar)}=\dfrac{b}{a}\]
The slope of \[BC\] is given as
\[\Rightarrow \dfrac{(br-b{{r}^{2}})}{(ar-a{{r}^{2}})}=\dfrac{b}{a}\]
We can clearly see that the slopes are equal and have \[B\] as common points. So, we can conclude that points \[A\] , \[B\] and \[C\] are collinear.
Therefore, \[A({{x}_{1}},{{y}_{1}})\] , \[B({{x}_{2}},{{y}_{2}})\] and \[C({{x}_{3}},{{y}_{3}})\] lie on the straight line.
Hence, option \[C\] is the correct answer.
So, the correct answer is “Option C”.
Note: Geometric progression problems require knowledge of exponent properties. Exponent properties are also called laws of indices. Exponent is the power of the base value. Power is the expression that represents repeated multiplication of the same number. Exponent is the quantity representing the power to which the number is raised.
\[{{n}^{th}}term=a{{r}^{n-1}}\]
The sum of \[n\] terms is given by \[\dfrac{a(1-{{r}^{n}})}{(1-r)}\] , when \[r<1\] and \[\dfrac{a({{r}^{n}}-1)}{(r-1)}\] , when \[r>1\] .
Geometric progression is the series of non-zero numbers in which each term after the first is found by multiplying the previous number by a fixed non-zero number called the common ratio.
If all the terms of G.P. are multiplied or divided by the same non-zero constant, then it remains a G.P. with the same common ratio. The geometric mean of \[a\] and \[b\] are given by \[\sqrt{ab}\] . The reciprocals of the terms of a given G.P. form a G.P. If each term of a G.P. is raised to the same power the resulting sequence also forms a G.P.
Complete step-by-step answer:
Given that \[{{x}_{1}},{{x}_{2}},{{x}_{3}}\] and \[{{y}_{1}},{{y}_{2,}}{{y}_{3}}\] are in G.P. with the same common ratio.
Let us assume that \[{{x}_{1}}=a\] , \[{{x}_{2}}=ar\] , \[{{x}_{3}}=a{{r}^{2}}\]
And \[{{y}_{1}}=b\] , \[{{y}_{2}}=br\] , \[{{y}_{3}}=b{{r}^{2}}\]
So, \[A({{x}_{1}},{{y}_{1}})\] , \[B({{x}_{2}},{{y}_{2}})\] and \[C({{x}_{3}},{{y}_{3}})\] can be written as \[A(a,b)\] , \[B(ar,br)\] and \[C(a{{r}^{2}},b{{r}^{2}})\]
Now we need to calculate the slope
The slope of \[AB\] is given as
\[\Rightarrow \dfrac{(b-br)}{(a-ar)}=\dfrac{b}{a}\]
The slope of \[BC\] is given as
\[\Rightarrow \dfrac{(br-b{{r}^{2}})}{(ar-a{{r}^{2}})}=\dfrac{b}{a}\]
We can clearly see that the slopes are equal and have \[B\] as common points. So, we can conclude that points \[A\] , \[B\] and \[C\] are collinear.
Therefore, \[A({{x}_{1}},{{y}_{1}})\] , \[B({{x}_{2}},{{y}_{2}})\] and \[C({{x}_{3}},{{y}_{3}})\] lie on the straight line.
Hence, option \[C\] is the correct answer.
So, the correct answer is “Option C”.
Note: Geometric progression problems require knowledge of exponent properties. Exponent properties are also called laws of indices. Exponent is the power of the base value. Power is the expression that represents repeated multiplication of the same number. Exponent is the quantity representing the power to which the number is raised.
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