Question

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

Answer 1

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.

Complete step by step answer:
Let \[A\] denote the first term and \[R\] denote the common ratio.
The \[{{n}^{th}}\] term has to be calculated by using the formula \[{{n}^{th}}term=A{{R}^{n-1}}\].
The \[{{n}^{th}}\] term,
\[{{A}_{n}}=A{{R}^{n-1}}\]
According to the question, the \[{{p}^{th}}\] term can be expressed as,
\[{{p}^{th}}term=A{{R}^{p-1}}\]
The \[{{q}^{th}}\] term can be expressed as
\[{{q}^{th}}term=A{{R}^{q-1}}\]
The \[{{r}^{th}}\] term can be expressed as
\[{{r}^{th}}term=A{{R}^{r-1}}\]
The \[{{p}^{th}}\] term is equal to \[a\]. So we can rewrite the equation as
\[a=A{{R}^{p-1}}\]

The \[{{q}^{th}}\] term is equal to \[b\]. So we can rewrite the equation as
\[b=A{{R}^{q-1}}\]
The \[{{r}^{th}}\] term is equal to \[c\]. So we can rewrite the equation as
\[c=A{{R}^{r-1}}\]
Substituting the equation values \[2\].
Using the exponent properties to solve the equation
\[{{a}^{q-r}}\cdot {{b}^{r-p}}\cdot {{c}^{p-q}}={{A}^{q-r}}{{R}^{(p-1)(}}^{q-r)}.{{A}^{r-p}}{{R}^{(q-1)(}}^{r-p)}.{{A}^{p-q}}{{R}^{(r-1)(}}^{p-q)}\]
Exponent property used in above equation
\[{{({{x}^{m}})}^{n}}={{x}^{m\times n}}\]

Again using the exponent properties
\[{{a}^{q-r}}\cdot {{b}^{r-p}}\cdot {{c}^{p-q}}={{A}^{(q-r)+(r-p)+(p-q)}}{{R}^{(p-1)(}}{{^{q-r)+}}^{(q-1)(}}{{^{r-p)+}}^{(r-1)(}}^{p-q)}\]
Exponent property used in above equation
\[{{x}^{m}}\times {{x}^{n}}={{x}^{m+n}}\]
On simplifying the solution we get
\[{{a}^{q-r}}\cdot {{b}^{r-p}}\cdot {{c}^{p-q}}={{A}^{0}}{{R}^{pr-pr-q+r+qr-pq-r+p+pr-qr-p+q}}\]
On further solving we get
\[{{a}^{q-r}}\cdot {{b}^{r-p}}\cdot {{c}^{p-q}}={{A}^{0}}{{R}^{0}}\]
Applying exponent property \[{{x}^{0}}=1\] we get
\[{{a}^{q-r}}\cdot {{b}^{r-p}}\cdot {{c}^{p-q}}=1\]

Therefore option B is the correct answer.

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.