In a arithmetic progression of 16 distinct terms with `a_(1)=16`, the sum is equal to square of the last term. The common difference of the A.P. is

Options

A. `8//15`
B. `-4//5`
C. `-8//5`
D. `-8//15`

Answer

To solve the problem step by step, we will follow these steps:

### Step 1: Define the terms of the arithmetic progression (A.P.)
Let the first term \( a_1 = 16 \) and the common difference be \( d \). The terms of the A.P. can be expressed as:
- \( a_1 = 16 \)
- \( a_2 = 16 + d \)
- \( a_3 = 16 + 2d \)
- ...
- \( a_{16} = 16 + 15d \)

### Step 2: Calculate the last term
The last term \( a_{16} \) can be written as:
\[ a_{16} = 16 + 15d \]

### Step 3: Calculate the sum of the first 16 terms
The sum \( S_n \) of the first \( n \) terms of an A.P. is given by the formula:
\[ S_n = \frac{n}{2} \times (2a + (n-1)d) \]
For \( n = 16 \):
\[ S_{16} = \frac{16}{2} \times (2 \times 16 + (16 - 1)d) = 8 \times (32 + 15d) = 256 + 120d \]

### Step 4: Set up the equation based on the problem statement
According to the problem, the sum of the terms is equal to the square of the last term:
\[ S_{16} = (a_{16})^2 \]
Substituting the expressions we found:
\[ 256 + 120d = (16 + 15d)^2 \]

### Step 5: Expand the right side of the equation
Expanding \( (16 + 15d)^2 \):
\[ (16 + 15d)^2 = 256 + 2 \times 16 \times 15d + (15d)^2 = 256 + 480d + 225d^2 \]

### Step 6: Set the equation
Now we have:
\[ 256 + 120d = 256 + 480d + 225d^2 \]
Subtract \( 256 \) from both sides:
\[ 120d = 480d + 225d^2 \]

### Step 7: Rearrange the equation
Rearranging gives:
\[ 0 = 225d^2 + 480d - 120d \]
\[ 0 = 225d^2 + 360d \]

### Step 8: Factor out the common term
Factoring out \( d \):
\[ d(225d + 360) = 0 \]

### Step 9: Solve for \( d \)
This gives us two solutions:
1. \( d = 0 \)
2. \( 225d + 360 = 0 \)

Solving \( 225d + 360 = 0 \):
\[ 225d = -360 \implies d = -\frac{360}{225} = -\frac{8}{5} \]

### Conclusion
The common difference \( d \) of the arithmetic progression is:
\[ d = -\frac{8}{5} \]