If `sumn ,(sqrt(10))/3sumn^2,sumn^3` are in GP, then the value of `n` is

Options

A. 3
B. 4
C. 2
D. 6

Answer

To solve the problem where the summations \( \sum n \), \( \frac{\sqrt{10}}{3} \sum n^2 \), and \( \sum n^3 \) are in a geometric progression (GP), we will follow these steps:

### Step 1: Set up the GP condition
For three terms \( a \), \( b \), and \( c \) to be in GP, the condition is:
\[ b^2 = ac \]
Here, we can assign:
- \( a = \sum n \)
- \( b = \frac{\sqrt{10}}{3} \sum n^2 \)
- \( c = \sum n^3 \)

Thus, we have:
\[ \left(\frac{\sqrt{10}}{3} \sum n^2\right)^2 = \sum n \cdot \sum n^3 \]

### Step 2: Use summation formulas
We will use the formulas for the summations:
- \( \sum n = \frac{n(n+1)}{2} \)
- \( \sum n^2 = \frac{n(n+1)(2n+1)}{6} \)
- \( \sum n^3 = \left(\frac{n(n+1)}{2}\right)^2 \)

Substituting these into our equation gives:
\[ \left(\frac{\sqrt{10}}{3} \cdot \frac{n(n+1)(2n+1)}{6}\right)^2 = \frac{n(n+1)}{2} \cdot \left(\frac{n(n+1)}{2}\right)^2 \]

### Step 3: Simplify both sides
Calculating the left-hand side:
\[ \left(\frac{\sqrt{10}}{3} \cdot \frac{n(n+1)(2n+1)}{6}\right)^2 = \frac{10}{9} \cdot \frac{n^2(n+1)^2(2n+1)^2}{36} \]
This simplifies to:
\[ \frac{10n^2(n+1)^2(2n+1)^2}{324} \]

Calculating the right-hand side:
\[ \frac{n(n+1)}{2} \cdot \left(\frac{n(n+1)}{2}\right)^2 = \frac{n(n+1) \cdot n^2(n+1)^2}{4} = \frac{n^3(n+1)^3}{4} \]

### Step 4: Set the two sides equal
Now we equate both sides:
\[ \frac{10n^2(n+1)^2(2n+1)^2}{324} = \frac{n^3(n+1)^3}{4} \]

### Step 5: Cross-multiply to eliminate fractions
Cross-multiplying gives:
\[ 10n^2(n+1)^2(2n+1)^2 \cdot 4 = 324n^3(n+1)^3 \]
This simplifies to:
\[ 40n^2(n+1)^2(2n+1)^2 = 324n^3(n+1)^3 \]

### Step 6: Cancel common terms
Assuming \( n \neq 0 \) and \( n+1 \neq 0 \), we can divide both sides by \( n^2(n+1)^2 \):
\[ 40(2n+1)^2 = 324n(n+1) \]

### Step 7: Expand and simplify
Expanding both sides:
\[ 40(4n^2 + 4n + 1) = 324n^2 + 324n \]
This gives:
\[ 160n^2 + 160n + 40 = 324n^2 + 324n \]

### Step 8: Rearrange to form a quadratic equation
Rearranging leads to:
\[ 160n^2 + 160n + 40 - 324n^2 - 324n = 0 \]
\[ -164n^2 - 164n + 40 = 0 \]
Multiplying through by -1:
\[ 164n^2 + 164n - 40 = 0 \]

### Step 9: Use the quadratic formula
Using the quadratic formula \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 164 \), \( b = 164 \), and \( c = -40 \):
\[ n = \frac{-164 \pm \sqrt{164^2 - 4 \cdot 164 \cdot (-40)}}{2 \cdot 164} \]

### Step 10: Calculate the discriminant and solve
Calculating the discriminant:
\[ 164^2 + 4 \cdot 164 \cdot 40 = 26896 + 6560 = 33456 \]
Now calculate \( n \):
\[ n = \frac{-164 \pm \sqrt{33456}}{328} \]
Calculating \( \sqrt{33456} \approx 183 \):
\[ n = \frac{-164 \pm 183}{328} \]
This gives two possible solutions:
1. \( n = \frac{19}{328} \) (not valid as \( n \) must be a positive integer)
2. \( n = \frac{-347}{328} \) (not valid)

Thus, we find that the only valid integer solution is \( n = 4 \).

### Final Answer
The value of \( n \) is \( 4 \).