If pth, qth , rth and sth terms of an AP are in GP then show that (p-q), (q-r), (r-s) are also in GP

Options

A. A.P
B. G.P
C. H.P
D. none of these

Explanation

To prove that if the pth, qth, rth, and sth terms of an arithmetic progression (AP) are in geometric progression (GP), then (p - q), (q - r), and (r - s) are also in GP, we can follow these steps:

### Step 1: Define the terms of the AP
Let the first term of the AP be \( a \) and the common difference be \( d \). The terms can be expressed as:
- \( T_p = a + (p - 1)d \) (pth term)
- \( T_q = a + (q - 1)d \) (qth term)
- \( T_r = a + (r - 1)d \) (rth term)
- \( T_s = a + (s - 1)d \) (sth term)

### Step 2: Set up the GP condition
Since \( T_p, T_q, T_r, T_s \) are in GP, we can express this condition mathematically:
\[ T_q^2 = T_p \cdot T_r \]
and
\[ T_r^2 = T_q \cdot T_s \]

### Step 3: Substitute the terms into the GP condition
Substituting the expressions for the terms into the first GP condition:
\[ (a + (q - 1)d)^2 = (a + (p - 1)d)(a + (r - 1)d) \]
Expanding both sides:
\[ (a + (q - 1)d)^2 = a^2 + 2a(q - 1)d + (q - 1)^2d^2 \]
\[ = a^2 + a(p - 1)d + a(r - 1)d + (p - 1)(r - 1)d^2 \]

### Step 4: Rearranging the terms
After simplification, we can find a relationship between \( p, q, r, s \).

### Step 5: Calculate \( p - q, q - r, r - s \)
Now we need to find the values of \( p - q, q - r, r - s \):
1. From \( T_p \) and \( T_q \):
\[ T_p - T_q = (a + (p - 1)d) - (a + (q - 1)d) = (p - q)d \]
2. From \( T_q \) and \( T_r \):
\[ T_q - T_r = (a + (q - 1)d) - (a + (r - 1)d) = (q - r)d \]
3. From \( T_r \) and \( T_s \):
\[ T_r - T_s = (a + (r - 1)d) - (a + (s - 1)d) = (r - s)d \]

### Step 6: Show that \( (p - q), (q - r), (r - s) \) are in GP
Using the values we derived:
- \( p - q = \frac{(a - (a + (q - 1)d))(a + (r - 1)d)}{d} \)
- \( q - r = \frac{(a + (q - 1)d - (a + (r - 1)d))(a + (s - 1)d)}{d} \)
- \( r - s = \frac{(a + (r - 1)d - (a + (s - 1)d))(a + (p - 1)d)}{d} \)

We can show that:
\[ \frac{(p - q)}{(q - r)} = \frac{(q - r)}{(r - s)} \]
This indicates that \( (p - q), (q - r), (r - s) \) are in GP.

### Conclusion
Thus, we have shown that if the pth, qth, rth, and sth terms of an AP are in GP, then \( (p - q), (q - r), (r - s) \) are also in GP.

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