Suppose `(1)/(2)x, lx,+1l,lx-1l,1l` are in A.P., then sum to 10 terms of the A.P. is

Options

A. 54
B. 36
C. 28
D. none of these

Answer

To solve the problem, we need to find the value of \( x \) and \( l \) such that the terms \( \frac{x}{2}, lx, l, lx - l, 1 \) are in arithmetic progression (A.P.). Then, we will calculate the sum of the first 10 terms of the A.P.

### Step 1: Set up the equations for A.P.
For the terms to be in A.P., the difference between consecutive terms must be constant. Therefore, we can set up the following equations:

1. \( lx - \frac{x}{2} = l - lx \)
2. \( l - lx = (lx - l) - 1 \)

### Step 2: Simplify the equations
From the first equation:
\[ lx - \frac{x}{2} = l - lx \]
Rearranging gives:
\[ lx + lx = l + \frac{x}{2} \]
\[ 2lx = l + \frac{x}{2} \]
\[ 2lx - l = \frac{x}{2} \]
\[ l(2x - 1) = \frac{x}{2} \]

From the second equation:
\[ l - lx = (lx - l) - 1 \]
This simplifies to:
\[ l - lx = lx - l - 1 \]
Rearranging gives:
\[ 2l - 2lx = -1 \]
\[ 2l(1 - x) = -1 \]
\[ l = \frac{-1}{2(1 - x)} \]

### Step 3: Substitute \( l \) into the first equation
Substituting \( l \) from the second equation into the first gives:
\[ \frac{-1}{2(1 - x)}(2x - 1) = \frac{x}{2} \]
Cross-multiplying leads to:
\[ -1(2x - 1) = x(1 - x) \]
Expanding both sides:
\[ -2x + 1 = x - x^2 \]
Rearranging gives:
\[ x^2 - 3x + 1 = 0 \]

### Step 4: Solve the quadratic equation
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ x = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} \]
\[ x = \frac{3 \pm \sqrt{9 - 4}}{2} \]
\[ x = \frac{3 \pm \sqrt{5}}{2} \]

### Step 5: Find \( l \) using \( x \)
Now we can substitute \( x \) back into the equation for \( l \):
\[ l = \frac{-1}{2(1 - \frac{3 \pm \sqrt{5}}{2})} \]

### Step 6: Calculate the common difference \( d \)
The first term \( a = \frac{x}{2} \) and the common difference \( d \) can be calculated from the terms of A.P. using:
\[ d = lx - l \]

### Step 7: Calculate the sum of the first 10 terms
The sum of the first \( n \) terms of an A.P. is given by:
\[ S_n = \frac{n}{2} [2a + (n - 1)d] \]
Substituting \( n = 10 \), \( a = \frac{x}{2} \), and \( d \) gives us the final result.