To solve the problem, we need to analyze the consecutive binomial coefficients of the expansion of \((1+x)^n\) and determine the relationships between the expressions \(\frac{a+b}{a}\), \(\frac{b+c}{b}\), and \(\frac{c+d}{c}\).
### Step-by-Step Solution:
1. **Identify the Binomial Coefficients**:
Let \(a\), \(b\), \(c\), and \(d\) be the consecutive binomial coefficients of \((1+x)^n\):
- \(a = \binom{n}{r}\)
- \(b = \binom{n}{r+1}\)
- \(c = \binom{n}{r+2}\)
- \(d = \binom{n}{r+3}\)
2. **Calculate \(\frac{a+b}{a}\)**:
\[
\frac{a+b}{a} = \frac{\binom{n}{r} + \binom{n}{r+1}}{\binom{n}{r}} = 1 + \frac{\binom{n}{r+1}}{\binom{n}{r}}
\]
Using the property of binomial coefficients:
\[
\frac{\binom{n}{r+1}}{\binom{n}{r}} = \frac{n-r}{r+1}
\]
Therefore,
\[
\frac{a+b}{a} = 1 + \frac{n-r}{r+1}
\]
3. **Calculate \(\frac{b+c}{b}\)**:
\[
\frac{b+c}{b} = \frac{\binom{n}{r+1} + \binom{n}{r+2}}{\binom{n}{r+1}} = 1 + \frac{\binom{n}{r+2}}{\binom{n}{r+1}}
\]
Again using the property:
\[
\frac{\binom{n}{r+2}}{\binom{n}{r+1}} = \frac{n-r-1}{r+2}
\]
Thus,
\[
\frac{b+c}{b} = 1 + \frac{n-r-1}{r+2}
\]
4. **Calculate \(\frac{c+d}{c}\)**:
\[
\frac{c+d}{c} = \frac{\binom{n}{r+2} + \binom{n}{r+3}}{\binom{n}{r+2}} = 1 + \frac{\binom{n}{r+3}}{\binom{n}{r+2}}
\]
Using the property:
\[
\frac{\binom{n}{r+3}}{\binom{n}{r+2}} = \frac{n-r-2}{r+3}
\]
Therefore,
\[
\frac{c+d}{c} = 1 + \frac{n-r-2}{r+3}
\]
5. **Analyze the Results**:
We have:
- \(\frac{a+b}{a} = 1 + \frac{n-r}{r+1}\)
- \(\frac{b+c}{b} = 1 + \frac{n-r-1}{r+2}\)
- \(\frac{c+d}{c} = 1 + \frac{n-r-2}{r+3}\)
Let:
\[
x_1 = \frac{a+b}{a}, \quad x_2 = \frac{b+c}{b}, \quad x_3 = \frac{c+d}{c}
\]
We can see that \(x_1\), \(x_2\), and \(x_3\) are in Arithmetic Progression (AP) because the differences between consecutive terms are constant.
6. **Conclusion**:
Since \(x_1\), \(x_2\), and \(x_3\) are in AP, it follows that:
\[
\frac{a+b}{a}, \frac{b+c}{b}, \frac{c+d}{c} \text{ are in AP.}
\]
### Final Answer:
The expressions \(\frac{a+b}{a}\), \(\frac{b+c}{b}\), and \(\frac{c+d}{c}\) are in Arithmetic Progression (AP).
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