A sequence is called an A.P if the difference of a term and the previous term is always same i.e if `a_(n+1)- a_(n)= ` constant ( common difference ) for all `n in N `
For an A.P whose first term is 'a ' and common difference is d has `n^(th)` term as `t_n=a+(n-1)d`
Sum of n terms of an A.P. whose first term is a, last term is `l` and common difference is d is
`S_(n) = n/2 (2a +(n-1)d)=n/2 (a+a+(n-1)d)= n/2 (a+l)`
`a_1,a_2,a_3,....a_8` is an A.P. with common difference d. Then `a_(8-2k)-a_k(1 leklt4 k in N)` is equal to

Options

A. `(2k-8)d`
B. `(3k-8)d`
C. `(8-3k)d`
D. `(8-2k)d`

Answer

To solve the problem, we need to find the expression for \( a_{8-2k} - a_k \) in terms of the first term \( a \) and the common difference \( d \) of the arithmetic progression (A.P.).

### Step-by-Step Solution:

1. **Identify the Terms**:
The \( n^{th} \) term of an A.P. is given by the formula:
\[ a_n = a + (n-1)d \]
Here, \( a \) is the first term, \( d \) is the common difference, and \( n \) is the term number.

2. **Find \( a_{8-2k} \)**:
We substitute \( n = 8-2k \) into the formula for the \( n^{th} \) term:
\[ a_{8-2k} = a + (8-2k-1)d = a + (7-2k)d \]

3. **Find \( a_k \)**:
Now, we substitute \( n = k \) into the formula for the \( n^{th} \) term:
\[ a_k = a + (k-1)d \]

4. **Calculate \( a_{8-2k} - a_k \)**:
Now we need to find the difference:
\[ a_{8-2k} - a_k = (a + (7-2k)d) - (a + (k-1)d) \]
Simplifying this expression:
\[ a_{8-2k} - a_k = (7-2k)d - (k-1)d \]
\[ = 7d - 2kd - kd + d \]
\[ = 7d - 3kd \]

5. **Final Expression**:
Thus, the expression for \( a_{8-2k} - a_k \) is:
\[ a_{8-2k} - a_k = (7 - 3k)d \]

### Final Answer:
\[ a_{8-2k} - a_k = (7 - 3k)d \]