If `sum_(r=1)^(n) r(sqrt(10))/(3)sum_(r=1)^(n)r^(2),sum_(r=1)^(n) r^(3)` are in G.P., then the value of n, is

Options

A. 2
B. 3
C. 4
D. non-existent

Explanation

To solve the problem, we need to determine the value of \( n \) such that the expressions

\[ \sum_{r=1}^{n} r, \quad \frac{\sqrt{10}}{3} \sum_{r=1}^{n} r^2, \quad \sum_{r=1}^{n} r^3 \]

are in geometric progression (G.P.).

### Step-by-Step Solution:

1. **Understanding the condition for G.P.**:
For three terms \( a, b, c \) to be in G.P., the condition is:
\[ b^2 = ac \]
Here, let:
\[ a = \sum_{r=1}^{n} r, \quad b = \frac{\sqrt{10}}{3} \sum_{r=1}^{n} r^2, \quad c = \sum_{r=1}^{n} r^3 \]

2. **Calculating the sums**:
We know the formulas for the sums:
- The sum of the first \( n \) natural numbers:
\[ \sum_{r=1}^{n} r = \frac{n(n+1)}{2} \]
- The sum of the squares of the first \( n \) natural numbers:
\[ \sum_{r=1}^{n} r^2 = \frac{n(n+1)(2n+1)}{6} \]
- The sum of the cubes of the first \( n \) natural numbers:
\[ \sum_{r=1}^{n} r^3 = \left( \frac{n(n+1)}{2} \right)^2 \]

3. **Substituting the sums into the G.P. condition**:
Substitute the sums into the G.P. condition:
\[ \left( \frac{\sqrt{10}}{3} \cdot \frac{n(n+1)(2n+1)}{6} \right)^2 = \frac{n(n+1)}{2} \cdot \left( \frac{n(n+1)}{2} \right)^2 \]

4. **Simplifying the left side**:
The left side becomes:
\[ \frac{10}{9} \cdot \frac{n^2(n+1)^2(2n+1)^2}{36} \]

5. **Simplifying the right side**:
The right side becomes:
\[ \frac{n(n+1)}{2} \cdot \frac{n^2(n+1)^2}{4} = \frac{n^3(n+1)^3}{8} \]

6. **Setting the two sides equal**:
Now we set the two sides equal:
\[ \frac{10}{9} \cdot \frac{n^2(n+1)^2(2n+1)^2}{36} = \frac{n^3(n+1)^3}{8} \]

7. **Cross-multiplying to eliminate fractions**:
Cross-multiplying gives:
\[ 10 \cdot n^2(n+1)^2(2n+1)^2 = 36 \cdot \frac{9}{8} \cdot n^3(n+1)^3 \]

8. **Further simplification**:
This simplifies to:
\[ 80 \cdot n^2(n+1)^2(2n+1)^2 = 324 \cdot n^3(n+1)^3 \]

9. **Dividing both sides by \( n^2(n+1)^2 \)** (assuming \( n \neq 0 \)):
\[ 80(2n+1)^2 = 324n(n+1) \]

10. **Expanding and rearranging**:
Expanding gives:
\[ 80(4n^2 + 4n + 1) = 324n^2 + 324n \]
\[ 320n^2 + 320n + 80 = 324n^2 + 324n \]
Rearranging leads to:
\[ 4n^2 + 4n - 80 = 0 \]

11. **Solving the quadratic equation**:
Dividing by 4:
\[ n^2 + n - 20 = 0 \]
Using the quadratic formula:
\[ n = \frac{-1 \pm \sqrt{1 + 80}}{2} = \frac{-1 \pm 9}{2} \]
This gives:
\[ n = 4 \quad \text{or} \quad n = -5 \]

Since \( n \) must be a positive integer, we have:
\[ n = 4 \]

### Conclusion:
Thus, the value of \( n \) is \( 4 \).