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Question: If the \[{{p}^{th}}\] , \[{{q}^{th}}\] and \[{{r}^{th}}\] term of a GP are \[a,b,c\] respectively, t...

If the pth{{p}^{th}} , qth{{q}^{th}} and rth{{r}^{th}} term of a GP are a,b,ca,b,c respectively, then aq−r⋅br−p⋅cp−q{{a}^{q-r}}\cdot {{b}^{r-p}}\cdot {{c}^{p-q}} is equal to
A. 00
B. 11
C. abcabc
D. pqrpqr

Explanation

Solution

GP stands for Geometric Progression. a,ar,ar2,ar3,.....a,ar,a{{r}^{2}},a{{r}^{3}},..... are said to be in GP where first term is aa and common ratio is rr .The nth{{n}^{th}} term is given by nthterm=arn−1{{n}^{th}}term=a{{r}^{n-1}}. The sum of nn terms is given by a(1−rn)(1−r)\dfrac{a(1-{{r}^{n}})}{(1-r)}, when r<1r<1 and a(rn−1)(r−1)\dfrac{a({{r}^{n}}-1)}{(r-1)}, when r>1r>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 AA denote the first term and RR denote the common ratio.
The nth{{n}^{th}} term has to be calculated by using the formula nthterm=ARn−1{{n}^{th}}term=A{{R}^{n-1}}.
The nth{{n}^{th}} term,
An=ARn−1{{A}_{n}}=A{{R}^{n-1}}
According to the question, the pth{{p}^{th}} term can be expressed as,
pthterm=ARp−1{{p}^{th}}term=A{{R}^{p-1}}
The qth{{q}^{th}} term can be expressed as
qthterm=ARq−1{{q}^{th}}term=A{{R}^{q-1}}
The rth{{r}^{th}} term can be expressed as
rthterm=ARr−1{{r}^{th}}term=A{{R}^{r-1}}
The pth{{p}^{th}} term is equal to aa. So we can rewrite the equation as
a=ARp−1a=A{{R}^{p-1}}

The qth{{q}^{th}} term is equal to bb. So we can rewrite the equation as
b=ARq−1b=A{{R}^{q-1}}
The rth{{r}^{th}} term is equal to cc. So we can rewrite the equation as
c=ARr−1c=A{{R}^{r-1}}
Substituting the equation values 22.
Using the exponent properties to solve the equation
aq−r⋅br−p⋅cp−q=Aq−rR(p−1)(q−r).Ar−pR(q−1)(r−p).Ap−qR(r−1)(p−q){{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
(xm)n=xm×n{{({{x}^{m}})}^{n}}={{x}^{m\times n}}

Again using the exponent properties
aq−r⋅br−p⋅cp−q=A(q−r)+(r−p)+(p−q)R(p−1)(q−r)+(q−1)(r−p)+(r−1)(p−q){{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
xm×xn=xm+n{{x}^{m}}\times {{x}^{n}}={{x}^{m+n}}
On simplifying the solution we get
aq−r⋅br−p⋅cp−q=A0Rpr−pr−q+r+qr−pq−r+p+pr−qr−p+q{{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
aq−r⋅br−p⋅cp−q=A0R0{{a}^{q-r}}\cdot {{b}^{r-p}}\cdot {{c}^{p-q}}={{A}^{0}}{{R}^{0}}
Applying exponent property x0=1{{x}^{0}}=1 we get
aq−r⋅br−p⋅cp−q=1{{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.