Answer
Verified
40.2k+ 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.
Formula used:
GP where ‘a’ is the first word and r is the common ratio
\[{n^{th}}term = a{r^{n - 1}}\].
Complete step-by-step solution
General term of Geometric progression is
\[A{R^{n - 1}}\]
Given that, \[a,b,c\]are \[{p^{th}},{q^{th}}\]and\[{r^{th}}\] terms of a Geometric progression, then
\[{\rm{a}} = {\rm{A}}{{\rm{R}}^{{\rm{p}} - 1}}\]-- (1)
\[{\rm{b}} = {\rm{A}}{{\rm{R}}^{{\rm{q}} - 1}}\]-- (2)
\[{\rm{c}} = {\rm{A}}{{\rm{R}}^{{\rm{r}} - 1}}\]-- (3)
Therefore, according to the given question it becomes,
\[{\left( {\frac{{\rm{c}}}{{\rm{b}}}} \right)^{\rm{p}}}{\left( {\frac{{\rm{b}}}{{\rm{a}}}} \right)^{\rm{r}}}{\left( {\frac{{\rm{a}}}{{\rm{c}}}} \right)^{\rm{q}}}\]-- (4)
Now, we have to substitute the values from the equation (1), equation (2) and equation (3) in equation (4), we get
\[{\left( {\frac{{{\rm{A}}{{\rm{R}}^{{\rm{r}} - 1}}}}{{{\rm{A}}{{\rm{R}}^{\rm{q}}} - 1}}} \right)^{\rm{p}}}{\left( {\frac{{{\rm{A}}{{\rm{R}}^{q - 1}}}}{{{\rm{A}}{{\rm{R}}^{p - 1}}}}} \right)^{\rm{r}}}{\left( {\frac{{{\rm{A}}{{\rm{R}}^{{\rm{p}} - 1}}}}{{{\rm{A}}{{\rm{R}}^{{\rm{r}} - 1}}}}} \right)^{\rm{q}}}\]
Let us cancel the similar terms, so that we can get
\[ \Rightarrow {\left( {{{\rm{R}}^{{\rm{r}} - {\rm{q}}}}} \right)^{\rm{p}}}{\left( {{{\rm{R}}^{{\rm{q}} - {\rm{p}}}}} \right)^{\rm{r}}}{\left( {{{\rm{R}}^{{\rm{p}} - 1}}} \right)^{\rm{q}}}\]
On using the exponent properties to solve the equation, we get
\[ \Rightarrow {{\rm{R}}^{{\rm{pr}} - {\rm{pq}}q{\rm{qr}} - {\rm{pr}} - {\rm{pq}} - {\rm{q}}}}\]
Simplify the powers of the above equation, we get
\[ \Rightarrow {{\rm{R}}^0} = 1{\rm{ }}\]
Therefore, if \[a,b,c\] are \[{p^{th}},{q^{th}}\] and \[{r^{th}}\] terms of a Geometric progression, then\[{\left( {\frac{c}{b}} \right)^p}\left( {\frac{b}{a}} \right){\left( {\frac{a}{c}} \right)^q}\]is equal to \[1\].
Hence, the option A is correct.
Note:
Students are likely to make mistakes in these types of problems; exponent qualities must be understood in order to solve geometric progression problems. The laws of indices are another name for exponent characteristics. The power of the base value is the exponent. 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.
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.
Formula used:
GP where ‘a’ is the first word and r is the common ratio
\[{n^{th}}term = a{r^{n - 1}}\].
Complete step-by-step solution
General term of Geometric progression is
\[A{R^{n - 1}}\]
Given that, \[a,b,c\]are \[{p^{th}},{q^{th}}\]and\[{r^{th}}\] terms of a Geometric progression, then
\[{\rm{a}} = {\rm{A}}{{\rm{R}}^{{\rm{p}} - 1}}\]-- (1)
\[{\rm{b}} = {\rm{A}}{{\rm{R}}^{{\rm{q}} - 1}}\]-- (2)
\[{\rm{c}} = {\rm{A}}{{\rm{R}}^{{\rm{r}} - 1}}\]-- (3)
Therefore, according to the given question it becomes,
\[{\left( {\frac{{\rm{c}}}{{\rm{b}}}} \right)^{\rm{p}}}{\left( {\frac{{\rm{b}}}{{\rm{a}}}} \right)^{\rm{r}}}{\left( {\frac{{\rm{a}}}{{\rm{c}}}} \right)^{\rm{q}}}\]-- (4)
Now, we have to substitute the values from the equation (1), equation (2) and equation (3) in equation (4), we get
\[{\left( {\frac{{{\rm{A}}{{\rm{R}}^{{\rm{r}} - 1}}}}{{{\rm{A}}{{\rm{R}}^{\rm{q}}} - 1}}} \right)^{\rm{p}}}{\left( {\frac{{{\rm{A}}{{\rm{R}}^{q - 1}}}}{{{\rm{A}}{{\rm{R}}^{p - 1}}}}} \right)^{\rm{r}}}{\left( {\frac{{{\rm{A}}{{\rm{R}}^{{\rm{p}} - 1}}}}{{{\rm{A}}{{\rm{R}}^{{\rm{r}} - 1}}}}} \right)^{\rm{q}}}\]
Let us cancel the similar terms, so that we can get
\[ \Rightarrow {\left( {{{\rm{R}}^{{\rm{r}} - {\rm{q}}}}} \right)^{\rm{p}}}{\left( {{{\rm{R}}^{{\rm{q}} - {\rm{p}}}}} \right)^{\rm{r}}}{\left( {{{\rm{R}}^{{\rm{p}} - 1}}} \right)^{\rm{q}}}\]
On using the exponent properties to solve the equation, we get
\[ \Rightarrow {{\rm{R}}^{{\rm{pr}} - {\rm{pq}}q{\rm{qr}} - {\rm{pr}} - {\rm{pq}} - {\rm{q}}}}\]
Simplify the powers of the above equation, we get
\[ \Rightarrow {{\rm{R}}^0} = 1{\rm{ }}\]
Therefore, if \[a,b,c\] are \[{p^{th}},{q^{th}}\] and \[{r^{th}}\] terms of a Geometric progression, then\[{\left( {\frac{c}{b}} \right)^p}\left( {\frac{b}{a}} \right){\left( {\frac{a}{c}} \right)^q}\]is equal to \[1\].
Hence, the option A is correct.
Note:
Students are likely to make mistakes in these types of problems; exponent qualities must be understood in order to solve geometric progression problems. The laws of indices are another name for exponent characteristics. The power of the base value is the exponent. 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.
Recently Updated Pages
Let gx 1 + x x and fx left beginarray20c 1x 0 0x 0 class 12 maths JEE_Main
The number of ways in which 5 boys and 3 girls can-class-12-maths-JEE_Main
Find dfracddxleft left sin x rightlog x right A left class 12 maths JEE_Main
Distance of the point x1y1z1from the line fracx x2l class 12 maths JEE_Main
In a box containing 100 eggs 10 eggs are rotten What class 12 maths JEE_Main
dfracddxex + 3log x A ex cdot x2x + 3 B ex cdot xx class 12 maths JEE_Main
Other Pages
Which of the following is not a redox reaction A CaCO3 class 11 chemistry JEE_Main
If a wire of resistance R is stretched to double of class 12 physics JEE_Main
How many grams of concentrated nitric acid solution class 11 chemistry JEE_Main
Formula for number of images formed by two plane mirrors class 12 physics JEE_Main
10g of hydrogen and 64g of oxygen were filled in a class 11 chemistry JEE_Main
In order to convert Aniline into chlorobenzene the class 12 chemistry JEE_Main