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If the \[{p^{th}},{q^{th}}\] and \[{r^{th}}\] term of a G.P. are \[a,b,c\] respectively, then \[{a^{q - r}} \cdot {b^{r - p}} \cdot {c^{p - q}}\] is equal to
A. 0
B. 1

Last updated date: 09th Apr 2024
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MVSAT 2024
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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.
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.