Suppose p is the first of `n(ngt1)` arithmetic means between two positive numbers a and b and q the first of n harmonic means between the same two numbers.
The value of p is

Options

A. `(na+b)/(n+1)`
B. `(nb+a)/(n+1)`
C. `(na-b)/(n+1)`
D. `(nb-a)/(n+1)`

Explanation

To find the value of \( p \), which is the first of \( n \) arithmetic means between two positive numbers \( a \) and \( b \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Arithmetic Means**:
- When we insert \( n \) arithmetic means between \( a \) and \( b \), we have the sequence:
\[ a, a_1, a_2, \ldots, a_n, b \]
- Here, \( a_1 \) is the first arithmetic mean, which we denote as \( p \).

2. **Common Difference**:
- The common difference \( d \) between consecutive terms in an arithmetic sequence can be calculated using the formula:
\[ d = \frac{b - a}{n + 1} \]
- This formula arises because there are \( n + 1 \) intervals created by \( n \) means between \( a \) and \( b \).

3. **Finding \( p \)**:
- The first arithmetic mean \( p \) can be expressed as:
\[ p = a + d \]
- Substituting the value of \( d \):
\[ p = a + \frac{b - a}{n + 1} \]

4. **Simplifying the Expression**:
- To combine the terms, we can rewrite \( p \):
\[ p = a + \frac{b - a}{n + 1} = \frac{(n + 1)a + (b - a)}{n + 1} \]
- This simplifies to:
\[ p = \frac{na + b}{n + 1} \]

### Final Answer:
Thus, the value of \( p \) is:
\[ p = \frac{na + b}{n + 1} \]