All roots of equation `x ^(5) -40 x ^(4) + alphax ^(3) + beta x ^(2) + gamma x + delta =0` are in G.P. if the sum of their reciprocals is 10, then `delta` can be equal to :

Options

A. 32
B. `-32`
C. `1/32`
D. `-1/32`

Answer

To solve the problem step by step, we will analyze the given polynomial and the conditions provided.

### Step 1: Understanding the Roots in G.P.
Let the roots of the polynomial \( x^5 - 40x^4 + \alpha x^3 + \beta x^2 + \gamma x + \delta = 0 \) be \( a, ar, ar^2, ar^3, ar^4 \), where \( a \) is the first term and \( r \) is the common ratio of the geometric progression (G.P.).

### Step 2: Sum of Reciprocals
We are given that the sum of the reciprocals of the roots is 10:
\[ \frac{1}{a} + \frac{1}{ar} + \frac{1}{ar^2} + \frac{1}{ar^3} + \frac{1}{ar^4} = 10 \]
This can be simplified as:
\[ \frac{1}{a} \left(1 + \frac{1}{r} + \frac{1}{r^2} + \frac{1}{r^3} + \frac{1}{r^4}\right) = 10 \]
The term inside the parentheses is a geometric series with first term 1 and common ratio \( \frac{1}{r} \). The sum of this series is:
\[ \frac{1 - \left(\frac{1}{r}\right)^5}{1 - \frac{1}{r}} = \frac{r^5 - 1}{r^5(r - 1)} \]
Thus, we have:
\[ \frac{1}{a} \cdot \frac{r^5 - 1}{r^5(r - 1)} = 10 \]
This leads to:
\[ \frac{r^5 - 1}{a \cdot r^5 (r - 1)} = 10 \]

### Step 3: Sum of Roots
Using Vieta's formulas, the sum of the roots \( a + ar + ar^2 + ar^3 + ar^4 \) is equal to 40 (the coefficient of \( x^4 \) with a negative sign):
\[ a(1 + r + r^2 + r^3 + r^4) = 40 \]
The sum \( 1 + r + r^2 + r^3 + r^4 \) is also a geometric series:
\[ \frac{r^5 - 1}{r - 1} \]
Thus, we can write:
\[ a \cdot \frac{r^5 - 1}{r - 1} = 40 \]

### Step 4: Relating the Two Equations
From the two equations derived:
1. \( \frac{r^5 - 1}{a \cdot r^5 (r - 1)} = 10 \)
2. \( a \cdot \frac{r^5 - 1}{r - 1} = 40 \)

From the second equation, we can express \( a \):
\[ a = \frac{40(r - 1)}{r^5 - 1} \]
Substituting this into the first equation gives:
\[ \frac{r^5 - 1}{\frac{40(r - 1)}{r^5 - 1} \cdot r^5 (r - 1)} = 10 \]
This simplifies to:
\[ \frac{(r^5 - 1)^2}{40r^5(r - 1)^2} = 10 \]
Multiplying both sides by \( 40r^5(r - 1)^2 \):
\[ (r^5 - 1)^2 = 400r^5(r - 1)^2 \]

### Step 5: Finding Delta
The product of the roots \( \delta \) is given by:
\[ \delta = a \cdot (ar) \cdot (ar^2) \cdot (ar^3) \cdot (ar^4) = a^5 \cdot r^{10} \]
From the earlier derived relationship \( ar^2 = \pm 2 \):
\[ \delta = (ar^2)^5 = (\pm 2)^5 = \pm 32 \]
Thus, \( \delta \) can be equal to \( 32 \) or \( -32 \).

### Final Answer
\[ \delta = 32 \text{ or } \delta = -32 \]