Given that `r_(n-1)+r_(n)=r_(n+1)` where r represents Bohr's orbit for `(n-1)^("th"),n^("th")` and `(n+1)^("th")` orbit, then n is

Answer

To solve the problem, we start with the given equation:

\[ r_{n-1} + r_n = r_{n+1} \]

### Step 1: Understand the relationship of the radius in Bohr's model
In Bohr's model of the atom, the radius of the nth orbit is given by the formula:

\[ r_n \propto \frac{n^2}{Z} \]

where \( Z \) is the atomic number. For simplicity, we can express the radius as:

\[ r_n = k \cdot n^2 \]

where \( k \) is a constant that depends on \( Z \).

### Step 2: Write the expressions for \( r_{n-1} \), \( r_n \), and \( r_{n+1} \)
Using the formula for the radius, we can write:

- \( r_{n-1} = k \cdot (n-1)^2 \)
- \( r_n = k \cdot n^2 \)
- \( r_{n+1} = k \cdot (n+1)^2 \)

### Step 3: Substitute these expressions into the equation
Substituting these expressions into the given equation:

\[ k \cdot (n-1)^2 + k \cdot n^2 = k \cdot (n+1)^2 \]

### Step 4: Simplify the equation
We can factor out \( k \) (assuming \( k \neq 0 \)):

\[ (n-1)^2 + n^2 = (n+1)^2 \]

Expanding each term:

- Left side:
\[ (n-1)^2 = n^2 - 2n + 1 \]
\[ n^2 = n^2 \]
So,
\[ (n-1)^2 + n^2 = n^2 - 2n + 1 + n^2 = 2n^2 - 2n + 1 \]

- Right side:
\[ (n+1)^2 = n^2 + 2n + 1 \]

Now we have:

\[ 2n^2 - 2n + 1 = n^2 + 2n + 1 \]

### Step 5: Rearranging the equation
Rearranging gives:

\[ 2n^2 - 2n + 1 - n^2 - 2n - 1 = 0 \]

This simplifies to:

\[ n^2 - 4n = 0 \]

### Step 6: Factor the quadratic equation
Factoring out \( n \):

\[ n(n - 4) = 0 \]

This gives us two solutions:

1. \( n = 0 \)
2. \( n = 4 \)

Since \( n \) represents the principal quantum number of an orbit, we discard \( n = 0 \) as it does not correspond to a physical orbit.

### Final Answer
Thus, the value of \( n \) is:

\[ n = 4 \]