To solve the problem, we need to find the ratio \( \frac{a_q}{a_p} \) given that \( a_1, a_2, a_3, \ldots \) are in Arithmetic Progression (A.P.) and \( a_p, a_q, a_r \) are in Geometric Progression (G.P.).
### Step 1: Express the terms in A.P.
Since \( a_1, a_2, a_3, \ldots \) are in A.P., we can express the terms as:
- \( a_1 = a \)
- \( a_2 = a + d \)
- \( a_3 = a + 2d \)
- In general, \( a_n = a + (n-1)d \)
### Step 2: Write the terms in G.P.
Given that \( a_p, a_q, a_r \) are in G.P., we can express these terms using the A.P. representation:
- \( a_p = a + (p-1)d \)
- \( a_q = a + (q-1)d \)
- \( a_r = a + (r-1)d \)
### Step 3: Use the property of G.P.
For three numbers to be in G.P., the square of the middle term must be equal to the product of the other two terms:
\[
a_q^2 = a_p \cdot a_r
\]
### Step 4: Substitute the A.P. expressions into the G.P. condition
Substituting the expressions we have:
\[
(a + (q-1)d)^2 = (a + (p-1)d)(a + (r-1)d)
\]
### Step 5: Expand both sides
Expanding the left side:
\[
(a + (q-1)d)^2 = a^2 + 2a(q-1)d + (q-1)^2d^2
\]
Expanding the right side:
\[
(a + (p-1)d)(a + (r-1)d) = a^2 + (p-1 + r-1)ad + (p-1)(r-1)d^2
\]
### Step 6: Set the expanded forms equal to each other
Equating both expansions:
\[
a^2 + 2a(q-1)d + (q-1)^2d^2 = a^2 + (p + r - 2)ad + (p-1)(r-1)d^2
\]
### Step 7: Simplify the equation
Cancel \( a^2 \) from both sides:
\[
2a(q-1)d + (q-1)^2d^2 = (p + r - 2)ad + (p-1)(r-1)d^2
\]
### Step 8: Rearranging the equation
Rearranging gives us:
\[
(2(q-1) - (p + r - 2))ad + ((q-1)^2 - (p-1)(r-1))d^2 = 0
\]
### Step 9: Solve for the ratio \( \frac{a_q}{a_p} \)
From the G.P. condition, we can derive:
\[
\frac{a_q}{a_p} = \frac{a + (q-1)d}{a + (p-1)d}
\]
This can be simplified further based on the values of \( p \), \( q \), and \( r \).
### Final Result
The ratio \( \frac{a_q}{a_p} \) can be expressed in terms of \( p \), \( q \), and \( r \) as:
\[
\frac{a_q}{a_p} = \sqrt{\frac{a_r}{a_p}}
\]
