Suppose `a,b,c,gt0` and a,b,c are the pth, qth, rth terms of a G.P. Let
`Delta=|(1,p,loga),(1,q,logb),(1,r,logc)|`
the numerical value of `Delta` is

Options

A. `-1`
B. `2`
C. `0`
D. None of these

Answer

To solve the problem, we need to calculate the determinant \( \Delta \) given by:

\[ \Delta = \begin{vmatrix} 1 & p & \log a \\ 1 & q & \log b \\ 1 & r & \log c \end{vmatrix} \]

### Step 1: Set up the determinant

We start with the determinant as defined:

\[ \Delta = \begin{vmatrix} 1 & p & \log a \\ 1 & q & \log b \\ 1 & r & \log c \end{vmatrix} \]

### Step 2: Apply row operations

We can simplify the determinant by performing row operations. We will replace the second and third rows by subtracting the first row from them:

\[ R_2 \leftarrow R_2 - R_1 \quad \text{and} \quad R_3 \leftarrow R_3 - R_1 \]

This gives us:

\[ \Delta = \begin{vmatrix} 1 & p & \log a \\ 0 & q-p & \log b - \log a \\ 0 & r-p & \log c - \log a \end{vmatrix} \]

### Step 3: Factor out the first column

Since the first column has a common factor of 1, we can factor it out:

\[ \Delta = 1 \cdot \begin{vmatrix} q-p & \log b - \log a \\ r-p & \log c - \log a \end{vmatrix} \]

### Step 4: Calculate the 2x2 determinant

Now we calculate the 2x2 determinant:

\[ \Delta = (q-p)(\log c - \log a) - (r-p)(\log b - \log a) \]

### Step 5: Simplify using properties of logarithms

Using the properties of logarithms, we can rewrite the logarithmic differences:

\[ \Delta = (q-p) \log \left(\frac{c}{a}\right) - (r-p) \log \left(\frac{b}{a}\right) \]

### Step 6: Factor out common terms

We can factor out the common term \( \log \left(\frac{1}{a}\right) \):

\[ \Delta = \log \left(\frac{c^{q-p}}{b^{r-p}}\right) \]

### Step 7: Final simplification

Using the property of logarithms, we can express the final result as:

\[ \Delta = \log \left(\frac{c^{q-p}}{b^{r-p}}\right) = 0 \]

This implies that the numerical value of \( \Delta \) is:

\[ \Delta = 0 \]

### Final Answer:

The numerical value of \( \Delta \) is \( 0 \).

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