Consider the arithmetic progression $$3, 7, 11, \dots$$, and let $$A_n$$ denote the sum of the first $$n$$ terms of this progression. Then, calculate the value of $$\frac{1}{25} \sum_{n=1}^{25} A_n$$.

Question

Accepted Answer

1. **Formula for Sum of First $$n$$ Terms of AP ($$A_n$$):**  
   The sum of the first $$n$$ terms of an arithmetic progression is given by:  
   $$
   A_n = \frac{n}{2} \left(2a + (n-1)dight)
   $$  
   Here:  
   - $$a = 3$$ (first term),  
   - $$d = 4$$ (common difference).

Substituting these values:  
   $$
   A_n = \frac{n}{2} \left(6 + (n-1)4ight)
   $$  
   Simplify:  
   $$
   A_n = \frac{n}{2} \left(6 + 4n - 4ight) = \frac{n}{2} (4n + 2)
   $$  
   Further simplify:  
   $$
   A_n = n(2n + 1)
   $$

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2. **Expression for Summation ($$\sum_{n=1}^{25} A_n$$):**  
   Substituting $$A_n = n(2n + 1)$$:  
   $$
   \sum_{n=1}^{25} A_n = \sum_{n=1}^{25} n(2n + 1)
   $$  
   Expand the summation:  
   $$
   \sum_{n=1}^{25} A_n = 2\sum_{n=1}^{25} n^2 + \sum_{n=1}^{25} n
   $$

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3. **Use Summation Formulas:**  
   - Sum of squares:  
     $$
     \sum_{n=1}^{25} n^2 = \frac{25(25+1)(2	imes25+1)}{6} = \frac{25(26)(51)}{6} = 5525
     $$

- Sum of first $$n$$ natural numbers:  
     $$
     \sum_{n=1}^{25} n = \frac{25(25+1)}{2} = 325
     $$

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4. **Substitute Values into the Expression:**  
   Substitute into the summation formula:  
   $$
   \sum_{n=1}^{25} A_n = 2(5525) + 325 = 11050 + 325 = 11375
   $$

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5. **Calculate Final Value ($$\frac{1}{25} \sum_{n=1}^{25} A_n$$):**  
   Divide by $$25$$:  
   $$
   \frac{1}{25} \sum_{n=1}^{25} A_n = \frac{11375}{25} = 455
   $$

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Thus, the value of $$\frac{1}{25} \sum_{n=1}^{25} A_n$$ is **455**, and the correct answer is **C)**.

Answer

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