If \[a,b,c\] are \[{p^{th}},{q^{th}}\] and \[{r^{th}}\] terms of a G.P., then \[{\left( {\frac{c}{b}} \right)^p}\left( {\frac{b}{a}^r} \right){\left( {\frac{a}{c}} \right)^q}\] is equal to
A. 1
B. \[{a^P}{b^q}{c^r}\]
C. \[{a^q}{b^r}{c^p}\]
D. \[{a^r}{b^p}{c^q}\]
Answer
108.9k+ 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.
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