Answer
64.8k+ views
Hint:
Geometric Progression is abbreviated as GP. Th series \[a,ar,a{r^2},a{r^3},......\] are said to be in GP where ‘a’ is the first word and r is the common ratio. The n-th term is given as \[{n^{th}}term = a{r^{n - 1}}\]. Geometric progression is a non-zero number series in which each term following the first is determined by multiplying the preceding value by a fixed non-zero number known as the common ratio.
Complete step-by-step solution
We have been given that,
The \[{p^{th}},{q^{th}}\]\[{r^{th}}\] terms of a G.P. are \[a,b,c\] respectively.
We can have,
A as first term, r as common ratio
By using the formula \[{n^{{\rm{th }}}}\] term \[ = A{R^{n - 1}}\], the \[{n^{{\rm{th }}}}\]term has to be calculated:
The\[{n^{{\rm{th }}}}\]term,
\[{A_n} = A{R^{n - 1}}\]
The \[{p^{{\rm{th }}}}\] term can be expressed as the below, according to the question,
\[{p^{{\rm{th }}}}{\rm{ term }} = A{R^{p - 1}}\]
The \[{q^{{\rm{th }}}}\] term is,
\[{q^{{\rm{th }}}}{\rm{ term }} = A{R^{q - 1}}\]
The \[{r^{{\rm{th }}}}\] term is,
\[{r^{{\rm{th }}}}{\rm{ term }} = A{R^{r - 1}}\]
As\[a\]is equal to the \[{p^{{\rm{th }}}}\] term, we can rewrite the equation as
\[a = A{R^{p - 1}}\]
The \[{q^{{\rm{th }}}}\] term is equal to \[b\]. The equation can be written as,
\[b = A{R^{q - 1}}\]
The\[{r^{{\rm{th }}}}\]term is equal to \[c\]. The equation can be written as,
\[c = A{R^{r - 1}}\]
Now, use the exponent property to solve the equation:
\[{a^{q - r}} \cdot {b^{r - p}} \cdot {c^{p - q}} = {A^{q - r}}{R^{(p - 1)(q - r)}} \cdot {A^{r - p}}{R^{(q - 1)\left( {^{( - p)}} \right.}} \cdot {A^{p - q}}{R^{(r - 1)(p - q)}}\]
In above equation, the exponent property used is
\[{\left( {{x^m}} \right)^n} = {x^{m \times n}}\]
Now, we have to use the exponent properties again,
\[{a^{q - r}} \cdot {b^{r - p}} \cdot {c^{p - q}} = {A^{(q - r) + (r - p) + (p - q)}}{R^{(p - 1)(q - r) + (q - 1)\left({{ r - p) + }(r - 1)(p - q)} \right.}}\]
The property used is
\[{x^m} \times {x^n} = {x^{m + n}}\]
We have to simplify the solution, we get
\[{a^{q - r}} \cdot {b^{r - p}} \cdot {c^{p - q}} = {A^0}{R^{pr - pr - q + r + qr - pq - r + p + pr - qr - p + q}}\]
We have to solve further until solution is obtained,
\[{a^{q - r}} \cdot {b^{r - p}} \cdot {c^{p - q}} = {A^0}{R^0}\]
Let’s apply exponent property \[{x^0} = 1\] we get
\[{a^{q - r}} \cdot {b^{r - p}} \cdot {c^{p - q}} = 1\]
Therefore, \[{a^{q - r}} \cdot {b^{r - p}} \cdot {c^{p - q}}\] is equal to\[1\].
Hence, option B is the correct answer.
Note:
While students are solving geometric progression problems, it necessitates an understanding of exponent characteristics. Exponent properties are sometimes known as indices laws. The exponent is the base value's power. The expression for repeated multiplication of the same number is power. The exponent is the amount that represents the power to which a number is raised.
Geometric Progression is abbreviated as GP. Th series \[a,ar,a{r^2},a{r^3},......\] are said to be in GP where ‘a’ is the first word and r is the common ratio. The n-th term is given as \[{n^{th}}term = a{r^{n - 1}}\]. Geometric progression is a non-zero number series in which each term following the first is determined by multiplying the preceding value by a fixed non-zero number known as the common ratio.
Complete step-by-step solution
We have been given that,
The \[{p^{th}},{q^{th}}\]\[{r^{th}}\] terms of a G.P. are \[a,b,c\] respectively.
We can have,
A as first term, r as common ratio
By using the formula \[{n^{{\rm{th }}}}\] term \[ = A{R^{n - 1}}\], the \[{n^{{\rm{th }}}}\]term has to be calculated:
The\[{n^{{\rm{th }}}}\]term,
\[{A_n} = A{R^{n - 1}}\]
The \[{p^{{\rm{th }}}}\] term can be expressed as the below, according to the question,
\[{p^{{\rm{th }}}}{\rm{ term }} = A{R^{p - 1}}\]
The \[{q^{{\rm{th }}}}\] term is,
\[{q^{{\rm{th }}}}{\rm{ term }} = A{R^{q - 1}}\]
The \[{r^{{\rm{th }}}}\] term is,
\[{r^{{\rm{th }}}}{\rm{ term }} = A{R^{r - 1}}\]
As\[a\]is equal to the \[{p^{{\rm{th }}}}\] term, we can rewrite the equation as
\[a = A{R^{p - 1}}\]
The \[{q^{{\rm{th }}}}\] term is equal to \[b\]. The equation can be written as,
\[b = A{R^{q - 1}}\]
The\[{r^{{\rm{th }}}}\]term is equal to \[c\]. The equation can be written as,
\[c = A{R^{r - 1}}\]
Now, use the exponent property to solve the equation:
\[{a^{q - r}} \cdot {b^{r - p}} \cdot {c^{p - q}} = {A^{q - r}}{R^{(p - 1)(q - r)}} \cdot {A^{r - p}}{R^{(q - 1)\left( {^{( - p)}} \right.}} \cdot {A^{p - q}}{R^{(r - 1)(p - q)}}\]
In above equation, the exponent property used is
\[{\left( {{x^m}} \right)^n} = {x^{m \times n}}\]
Now, we have to use the exponent properties again,
\[{a^{q - r}} \cdot {b^{r - p}} \cdot {c^{p - q}} = {A^{(q - r) + (r - p) + (p - q)}}{R^{(p - 1)(q - r) + (q - 1)\left({{ r - p) + }(r - 1)(p - q)} \right.}}\]
The property used is
\[{x^m} \times {x^n} = {x^{m + n}}\]
We have to simplify the solution, we get
\[{a^{q - r}} \cdot {b^{r - p}} \cdot {c^{p - q}} = {A^0}{R^{pr - pr - q + r + qr - pq - r + p + pr - qr - p + q}}\]
We have to solve further until solution is obtained,
\[{a^{q - r}} \cdot {b^{r - p}} \cdot {c^{p - q}} = {A^0}{R^0}\]
Let’s apply exponent property \[{x^0} = 1\] we get
\[{a^{q - r}} \cdot {b^{r - p}} \cdot {c^{p - q}} = 1\]
Therefore, \[{a^{q - r}} \cdot {b^{r - p}} \cdot {c^{p - q}}\] is equal to\[1\].
Hence, option B is the correct answer.
Note:
While students are solving geometric progression problems, it necessitates an understanding of exponent characteristics. Exponent properties are sometimes known as indices laws. The exponent is the base value's power. The expression for repeated multiplication of the same number is power. The exponent is the amount that represents the power to which a number is raised.
Recently Updated Pages
Write a composition in approximately 450 500 words class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Arrange the sentences P Q R between S1 and S5 such class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What is the common property of the oxides CONO and class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What happens when dilute hydrochloric acid is added class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
If four points A63B 35C4 2 and Dx3x are given in such class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
The area of square inscribed in a circle of diameter class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Other Pages
Excluding stoppages the speed of a bus is 54 kmph and class 11 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
A boat takes 2 hours to go 8 km and come back to a class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Electric field due to uniformly charged sphere class 12 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
According to classical free electron theory A There class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
In the ground state an element has 13 electrons in class 11 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Differentiate between homogeneous and heterogeneous class 12 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)