In a G.P the sum of the first and last terms is 66, the product of the second and the last but one is 128 nd the sum of the terms is 126 in any case, the difference of the least and greatest terms is

Options

A. 2
B. `3/2`
C. `5/2`
D. `-1/2`

Explanation

To solve the problem step by step, let's denote the first term of the G.P. as \( a \) and the common ratio as \( r \).

### Step 1: Set up the equations based on the problem statement

1. The sum of the first and last terms is given as:
\[ a + ar^{n-1} = 66 \quad \text{(1)} \]

2. The product of the second term and the last but one term is given as:
\[ ar \cdot ar^{n-2} = 128 \quad \text{(2)} \]
This simplifies to:
\[ a^2 r^{n-1} = 128 \quad \text{(3)} \]

3. The sum of all terms in the G.P. is given as:
\[ S_n = \frac{a(1 - r^n)}{1 - r} = 126 \quad \text{(4)} \]

### Step 2: Solve for \( ar^{n-1} \) from equation (1)

From equation (1):
\[ ar^{n-1} = 66 - a \quad \text{(5)} \]

### Step 3: Substitute equation (5) into equation (3)

Substituting equation (5) into equation (3):
\[ a^2 (66 - a) = 128 \]
This expands to:
\[ 66a^2 - a^3 = 128 \]
Rearranging gives:
\[ a^3 - 66a^2 + 128 = 0 \quad \text{(6)} \]

### Step 4: Solve the cubic equation (6)

To solve the cubic equation \( a^3 - 66a^2 + 128 = 0 \), we can use the Rational Root Theorem or synthetic division to find possible rational roots. Testing \( a = 64 \) and \( a = 2 \):

1. For \( a = 64 \):
\[ 64^3 - 66 \cdot 64^2 + 128 = 0 \quad \text{(True)} \]

2. For \( a = 2 \):
\[ 2^3 - 66 \cdot 2^2 + 128 = 0 \quad \text{(True)} \]

Thus, \( a = 64 \) or \( a = 2 \).

### Step 5: Find the corresponding \( r \)

Using \( a = 64 \):
From equation (5):
\[ ar^{n-1} = 66 - 64 = 2 \quad \Rightarrow \quad 64r^{n-1} = 2 \quad \Rightarrow \quad r^{n-1} = \frac{2}{64} = \frac{1}{32} \quad \text{(7)} \]

Using \( a = 2 \):
From equation (5):
\[ ar^{n-1} = 66 - 2 = 64 \quad \Rightarrow \quad 2r^{n-1} = 64 \quad \Rightarrow \quad r^{n-1} = \frac{64}{2} = 32 \quad \text{(8)} \]

### Step 6: Determine \( n \)

From equation (7):
\[ r^{n-1} = \frac{1}{32} = 2^{-5} \quad \Rightarrow \quad n - 1 = -5 \quad \Rightarrow \quad n = -4 \quad \text{(not valid)} \]

From equation (8):
\[ r^{n-1} = 32 = 2^5 \quad \Rightarrow \quad n - 1 = 5 \quad \Rightarrow \quad n = 6 \quad \text{(valid)} \]

### Step 7: Calculate the difference between the greatest and least terms

The greatest term is \( ar^{n-1} \) and the least term is \( a \):
\[ \text{Greatest term} = ar^{n-1} = 64 \quad \text{(from a = 64)} \]
\[ \text{Least term} = a = 2 \]
Thus, the difference is:
\[ \text{Difference} = 64 - 2 = 62 \]

### Final Answer
The difference between the least and greatest terms is \( \boxed{62} \).