If `(m+1)^(th),(n+1)^(th),(r+1)^(th)` terms of the A.P. a, a+d ,a+2d , …… are in G.P. while m,n,r are in H.P. , then `(d)/(a)` =

Options

A. `(1)/(n)`
B. `-(1)/(n)`
C. `-(2)/(n)`
D. `(2)/(n)`

Answer

To solve the problem, we need to find the value of \(\frac{d}{a}\) given that the \((m+1)^{th}\), \((n+1)^{th}\), and \((r+1)^{th}\) terms of an arithmetic progression (A.P.) are in geometric progression (G.P.) and that \(m\), \(n\), and \(r\) are in harmonic progression (H.P.).

### Step-by-Step Solution:

1. **Identify the Terms of the A.P.**:
The general term of an A.P. can be expressed as:
\[ T_k = a + (k-1)d \]
Therefore, we can write:
- \((m+1)^{th}\) term:
\[ T_{m+1} = a + md \]
- \((n+1)^{th}\) term:
\[ T_{n+1} = a + nd \]
- \((r+1)^{th}\) term:
\[ T_{r+1} = a + rd \]

2. **Condition for G.P.**:
The terms \(T_{m+1}\), \(T_{n+1}\), and \(T_{r+1}\) are in G.P. if:
\[ (T_{n+1})^2 = T_{m+1} \cdot T_{r+1} \]
Substituting the terms:
\[ (a + nd)^2 = (a + md)(a + rd) \]

3. **Expand Both Sides**:
Expanding the left-hand side:
\[ (a + nd)^2 = a^2 + 2and + n^2d^2 \]
Expanding the right-hand side:
\[ (a + md)(a + rd) = a^2 + (m+r)ad + mrd^2 \]

4. **Set the Equations Equal**:
Equating both sides gives:
\[ a^2 + 2and + n^2d^2 = a^2 + (m+r)ad + mrd^2 \]

5. **Simplify the Equation**:
Cancel \(a^2\) from both sides:
\[ 2and + n^2d^2 = (m+r)ad + mrd^2 \]
Rearranging gives:
\[ n^2d^2 - mrd^2 + (2an - (m+r)ad) = 0 \]
Factor out \(d^2\):
\[ d^2(n^2 - mr) + ad(2n - (m+r)) = 0 \]

6. **Condition for H.P.**:
Since \(m\), \(n\), and \(r\) are in H.P., we have:
\[ \frac{2}{n} = \frac{1}{m} + \frac{1}{r} \]
This can be rearranged to:
\[ 2mr = n(m+r) \]

7. **Substituting \(n\)**:
Substitute \(n = \frac{2mr}{m+r}\) into the equation derived from the G.P. condition:
\[ d^2\left(\left(\frac{2mr}{m+r}\right)^2 - mr\right) + ad\left(2\left(\frac{2mr}{m+r}\right) - (m+r)\right) = 0 \]

8. **Solve for \(\frac{d}{a}\)**:
After some algebraic manipulation, we can express \(\frac{d}{a}\) in terms of \(m\), \(n\), and \(r\):
\[ \frac{d}{a} = \frac{r + m - 2n}{n^2 - mr} \]
Substitute \(n = \frac{2mr}{m+r}\) and simplify to find:
\[ \frac{d}{a} = -\frac{2}{n} \]

### Final Result:
Thus, the value of \(\frac{d}{a}\) is:
\[ \frac{d}{a} = -\frac{2}{n} \]