Question

If the p^th, q^th and rth terms of a G.P. are a, b and c, respectively. Prove that \(a^{q-r} b^{r-p} c^{p-q}=1.\)

Answer 1

Let the given geometric progression (G.P.) have first term \( A \) and common ratio \( R \).

Then the general term of the G.P. is

\[ T_n = AR^{\,n-1}. \]

It is given that the

\[ p^{\text{th}} \text{ term } = a, \quad q^{\text{th}} \text{ term } = b, \quad r^{\text{th}} \text{ term } = c. \]

Therefore,

\[ AR^{p-1} = a \quad \text{… (1)} \]

\[ AR^{q-1} = b \quad \text{… (2)} \]

\[ AR^{r-1} = c \quad \text{… (3)} \]

Consider the expression

\[ a^{\,q-r} \, b^{\,r-p} \, c^{\,p-q}. \]

Substituting the values of \( a, b, c \) from equations (1), (2), and (3), we get

\[ (AR^{p-1})^{q-r} (AR^{q-1})^{r-p} (AR^{r-1})^{p-q}. \]

Separating the powers of \( A \) and \( R \):

\[ A^{(q-r)+(r-p)+(p-q)} \cdot R^{(p-1)(q-r)+(q-1)(r-p)+(r-1)(p-q)}. \]

Simplifying the exponent of \( A \):

\[ (q-r)+(r-p)+(p-q) = 0. \]

Hence,

\[ A^0 = 1. \]

Now simplify the exponent of \( R \):

\[ (p-1)(q-r)+(q-1)(r-p)+(r-1)(p-q). \]

Expanding each term:

\[ = pq - pr - q + r + qr - qp - r + p + rp - rq - p + q. \]

On simplification, all terms cancel out, giving

\[ 0. \]

Therefore,

\[ R^0 = 1. \]

Hence,

\[ a^{q-r} b^{r-p} c^{p-q} = 1. \]

Thus, the required result is proved.