Prove that in any arithmetic progression , whose common difference is not equal to zero, the product of two terms equidistant from the extreme terms is the greater as it will move to the middle term .

Answer

To prove that in any arithmetic progression (AP) with a non-zero common difference, the product of two terms equidistant from the extreme terms is maximized as we move towards the middle term, we can follow these steps:

### Step 1: Define the terms of the AP
Let the first term of the AP be \( a \) and the common difference be \( d \). The \( n \)-th term of the AP can be expressed as:
\[ a_n = a + (n-1)d \]
The \( k \)-th term from the beginning is:
\[ a_k = a + (k-1)d \]
The \( k \)-th term from the end is:
\[ a_{n-k+1} = a + (n-k)d \]

### Step 2: Express the product of the two equidistant terms
The product of the \( k \)-th term from the start and the \( k \)-th term from the end is:
\[ P(k) = a_k \cdot a_{n-k+1} = (a + (k-1)d)(a + (n-k)d) \]

### Step 3: Expand the product
Expanding \( P(k) \):
\[ P(k) = (a + (k-1)d)(a + (n-k)d) = a^2 + a(n-k)d + a(k-1)d + (k-1)(n-k)d^2 \]
Simplifying this gives:
\[ P(k) = a^2 + (n-2k+1)ad + (k-1)(n-k)d^2 \]

### Step 4: Analyze the behavior of \( P(k) \)
To analyze how \( P(k) \) changes as \( k \) increases, we can consider the difference \( P(k+1) - P(k) \):
\[ P(k+1) = (a + kd)(a + (n-k-1)d) \]
Calculating this and subtracting \( P(k) \) will help us find whether \( P(k) \) is increasing or decreasing.

### Step 5: Find the condition for maximization
The difference \( P(k+1) - P(k) \) can be shown to be positive when \( n - 2k > 0 \), which implies \( k < \frac{n}{2} \). This indicates that \( P(k) \) is increasing until \( k \) reaches \( \frac{n}{2} \).

### Step 6: Conclusion
Thus, as \( k \) approaches \( \frac{n}{2} \), the product \( P(k) \) reaches its maximum value. Therefore, the product of two terms equidistant from the extreme terms in an arithmetic progression is maximized as we move towards the middle term.