Which of the following can be terms (not necessarily consecutive) of any A.P.? a. 1,6,19 b. `sqrt(2),sqrt(50),sqrt(98)` c. `log2,log16 ,log128` d. `sqrt(2),sqrt(3),sqrt(7)`

Options

A. 1,6,19
B. `sqrt(2).sqrt(50),sqrt(98)`
C. log 2,log 16 , log128
D. `sqrt(2),sqrt(3),sqrt(7)`

Explanation

To determine which of the given sets of numbers can be terms of an Arithmetic Progression (A.P.), we will use the property that for three terms \(a\), \(b\), and \(c\) to be in A.P., the following condition must hold:

\[ \frac{b - a}{c - b} = \frac{q - p}{r - q} \]

Where \(p\), \(q\), and \(r\) are the positions of the terms in the A.P.

Now, let's analyze each option step by step:

### Option a: 1, 6, 19
1. Let \(a = 1\), \(b = 6\), and \(c = 19\).
2. Calculate \(c - b\) and \(b - a\):
- \(c - b = 19 - 6 = 13\)
- \(b - a = 6 - 1 = 5\)
3. Now, check if \(\frac{c - b}{b - a}\) is a rational number:
\[ \frac{c - b}{b - a} = \frac{13}{5} \]
This is a rational number.

### Option b: \(\sqrt{2}, \sqrt{50}, \sqrt{98}\)
1. Let \(a = \sqrt{2}\), \(b = \sqrt{50}\), and \(c = \sqrt{98}\).
2. Calculate \(c - b\) and \(b - a\):
- \(c - b = \sqrt{98} - \sqrt{50}\)
- \(b - a = \sqrt{50} - \sqrt{2}\)
3. We need to simplify:
- \(\sqrt{98} = 7\sqrt{2}\)
- \(\sqrt{50} = 5\sqrt{2}\)
- Thus, \(c - b = 7\sqrt{2} - 5\sqrt{2} = 2\sqrt{2}\)
- \(b - a = 5\sqrt{2} - \sqrt{2} = 4\sqrt{2}\)
4. Now check if \(\frac{c - b}{b - a}\) is a rational number:
\[ \frac{c - b}{b - a} = \frac{2\sqrt{2}}{4\sqrt{2}} = \frac{1}{2} \]
This is a rational number.

### Option c: \(\log_2, \log_{16}, \log_{128}\)
1. Let \(a = \log_2\), \(b = \log_{16}\), and \(c = \log_{128}\).
2. Calculate \(c - b\) and \(b - a\):
- \(c - b = \log_{128} - \log_{16}\)
- \(b - a = \log_{16} - \log_2\)
3. We can express these logarithms in terms of \(\log_2\):
- \(\log_{16} = \log_2(2^4) = 4\)
- \(\log_{128} = \log_2(2^7) = 7\)
- Thus, \(c - b = 7 - 4 = 3\)
- \(b - a = 4 - 1 = 3\)
4. Now check if \(\frac{c - b}{b - a}\) is a rational number:
\[ \frac{c - b}{b - a} = \frac{3}{3} = 1 \]
This is a rational number.

### Option d: \(\sqrt{2}, \sqrt{3}, \sqrt{7}\)
1. Let \(a = \sqrt{2}\), \(b = \sqrt{3}\), and \(c = \sqrt{7}\).
2. Calculate \(c - b\) and \(b - a\):
- \(c - b = \sqrt{7} - \sqrt{3}\)
- \(b - a = \sqrt{3} - \sqrt{2}\)
3. We need to check if \(\frac{c - b}{b - a}\) is a rational number:
- This requires checking if \(\sqrt{7} - \sqrt{3}\) and \(\sqrt{3} - \sqrt{2}\) can be expressed as a rational number.
4. It turns out that this expression does not simplify to a rational number, thus:
\[ \frac{c - b}{b - a} \text{ is not a rational number.} \]

### Conclusion:
The terms that can be in an A.P. are from options **a**, **b**, and **c**. Option **d** cannot be in an A.P.