Statement-1: If a, b, c are positive numbers in AP such that `(1)/(ab)+(1)/(bc)+(1)/(ca)=1,` then the least value of b is `sqrt(3)`.
Statement-2: `A.M.geG.M.`

Options

A. 1
B. 2
C. 3
D. 4

Explanation

To solve the problem, we need to verify the statements given in the problem and derive the least value of \( b \) when \( a, b, c \) are in arithmetic progression (AP) and satisfy the equation:

\[ \frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca} = 1 \]

### Step 1: Express \( a, b, c \) in terms of \( b \)

Since \( a, b, c \) are in AP, we can express them as:
- \( a = b - d \)
- \( b = b \)
- \( c = b + d \)

for some \( d > 0 \).

### Step 2: Substitute \( a, b, c \) into the equation

Substituting \( a, b, c \) into the equation gives us:

\[ \frac{1}{(b-d)b} + \frac{1}{b(b+d)} + \frac{1}{(b-d)(b+d)} = 1 \]

### Step 3: Find a common denominator

The common denominator for the left-hand side is:

\[ (b-d)b(b+d) \]

Thus, we rewrite the equation as:

\[ \frac{(b+d) + (b-d) + b}{(b-d)b(b+d)} = 1 \]

### Step 4: Simplify the numerator

The numerator simplifies to:

\[ (b+d) + (b-d) + b = 3b \]

So, we have:

\[ \frac{3b}{(b-d)b(b+d)} = 1 \]

### Step 5: Cross-multiply and simplify

Cross-multiplying gives:

\[ 3b = (b-d)b(b+d) \]

Expanding the right-hand side:

\[ 3b = b^3 - d^2b \]

### Step 6: Rearranging the equation

Rearranging gives us:

\[ b^3 - d^2b - 3b = 0 \]

This simplifies to:

\[ b^3 - (d^2 + 3)b = 0 \]

### Step 7: Factor out \( b \)

Factoring out \( b \):

\[ b(b^2 - (d^2 + 3)) = 0 \]

Since \( b > 0 \), we have:

\[ b^2 - (d^2 + 3) = 0 \]

### Step 8: Solve for \( b \)

This gives us:

\[ b^2 = d^2 + 3 \]

Taking the square root:

\[ b = \sqrt{d^2 + 3} \]

### Step 9: Find the minimum value of \( b \)

To find the minimum value of \( b \), note that \( d \) is a positive number. The minimum occurs when \( d = 0 \):

\[ b = \sqrt{0 + 3} = \sqrt{3} \]

### Conclusion

Thus, the least value of \( b \) is \( \sqrt{3} \).