Question

If $a,b,c$ are ${p^{th}},{q^{th}}$ and ${r^{th}}$ terms of GP, then \left| {\begin{array}{*{20}{c}} {\log a}&p;&1 \\\ {\log b}&q;&1 \\\ {\log c}&r;&1 \end{array}} \right| is equal to ?
A. 0
B. 1
C. $\log abc$
D. $pqr$

Answer 1

Let $A$ be the first term and $R$ be the common ratio of the given G.P. and hence write the values of $a,b,c$ in term of $p,q,r$. Then, apply the properties of log to simplify the terms of the given determinant. Next, apply elementary row transformations to evaluate the value of determinant.

Complete step-by-step answer:
The ${p^{th}},{q^{th}}$ and ${r^{th}}$ terms of GP are $a,b,c$
Let $A$ be the first term and $R$ be the common ratio of the given G.P.
Then, ${p^{th}}$ term of the sequence can be written as $a = A{R^{p - 1}}$
Similarly, ${q^{th}}$ can be written as $b = A{R^{q - 1}}$ and ${r^{th}}$ term is $c = A{R^{r - 1}}$
On substituting the value of $a,b,c$ in the given determinant, we will get,
\left| {\begin{array}{*{20}{c}} {\log A{R^{p - 1}}}&p;&1 \\\ {\log A{R^{q - 1}}}&q;&1 \\\ {\log A{R^{r - 1}}}&r;&1 \end{array}} \right|
Now, we will use the properties of log, that are$\log nm = \log n + \log m$ $\log {n^m} = m{\log _n}$ to simplify the terms in determinant.

{\log A + \left( {p - 1} \right)\log R}&p;&1 \\\ {\log A + \left( {q - 1} \right)\log R}&q;&1 \\\ {\log A + \left( {r - 1} \right)\log R}&r;&1 \end{array}} \right|$$ Apply elementary row operation, ${R_1} \to {R_1} - {R_2}$ and ${R_2} \to {R_2} - {R_3}$ $$\left| {\begin{array}{*{20}{c}} {\left( {p - 1 - q + 1} \right)\log R}&{p - q}&0 \\\ {\left( {q - 1 - r + 1} \right)\log R}&{q - r}&0 \\\ {\log A + \left( {r - 1} \right)\log R}&r;&1 \end{array}} \right| = \left| {\begin{array}{*{20}{c}} {\left( {p - q} \right)\log R}&{p - q}&0 \\\ {\left( {q - r} \right)\log R}&{q - r}&0 \\\ {\log A + \left( {r - 1} \right)\log R}&r;&1 \end{array}} \right|$$ Now, we will take $p - q$ and $$q - r$$ common from the first and second row respectively. $$\left( {p - q} \right)\left( {q - r} \right)\left| {\begin{array}{*{20}{c}} {\log R}&1&0 \\\ {\log R}&1&0 \\\ {\log A + \left( {r - 1} \right)\log R}&r;&1 \end{array}} \right|$$ Here, the first and second row are the same and if any two rows are the same then the determinant is zero. Therefore, the determinant will be equal to 0. **Hence, option A is correct.** **Note:** Determinant is a number corresponding to the given matrix. The determinant can be calculated along any row or column. If any two rows or any two columns are the same, that is they have the same values, then the determinant is zero.