Question

Consider the arithmetic progression 3,7,11,…and let \(A_n\) denote the sum of the first n terms of this progression.Then the value of \(\frac{1}{25}∑^{25}_{n=1}A_n\) is

• A. 404
• B. 442
• C. 455
• D. 415

Answer 1

The Correct Option is C

Given

Arithmetic Progression (AP): 3, 7, 11, ...
First term: \( a = 3 \)
Common difference: \( d = 7 - 3 = 4 \)

 Step 1: General Formula for the Sum of First \( n \) Terms

\[ S_n = \frac{n}{2} \left(2a + (n - 1)d \right) \] Substituting \( a = 3 \) and \( d = 4 \): \[ S_n = \frac{n}{2} \left(2 \times 3 + (n - 1) \times 4 \right) = \frac{n}{2}(6 + 4n - 4) = \frac{n}{2}(4n + 2) = n(2n + 1) \]

 Step 2: Required Expression

Find: \[ \frac{1}{25} \sum_{n=1}^{25} A_n \quad \text{where } A_n = S_n = n(2n + 1) \] So: \[ \sum_{n=1}^{25} n(2n + 1) = \sum_{n=1}^{25} (2n^2 + n) \] Split into two separate summations: \[ \sum_{n=1}^{25} 2n^2 + \sum_{n=1}^{25} n = 2 \sum_{n=1}^{25} n^2 + \sum_{n=1}^{25} n \]

 Step 3: Use Summation Formulas

Sum of squares of first \( n \) natural numbers: \[ \sum_{n=1}^{25} n^2 = \frac{25 \cdot 26 \cdot 51}{6} = 5525 \] Therefore: \[ 2 \sum_{n=1}^{25} n^2 = 2 \cdot 5525 = 11050 \] Sum of first \( n \) natural numbers: \[ \sum_{n=1}^{25} n = \frac{25 \cdot 26}{2} = 325 \] Adding both: \[ \sum_{n=1}^{25} n(2n + 1) = 11050 + 325 = 11375 \]

 Step 4: Final Answer

\[ \frac{1}{25} \sum_{n=1}^{25} A_n = \frac{11375}{25} = \boxed{455} \]

Therefore, the final answer is 455.

Answer 2

A.P. Information

Given an arithmetic progression (A.P.): \( 3, 7, 11, \dots \) 
First term: \( a = 3 \), common difference: \( d = 4 \)

 Step 1: General Sum Formula

Sum of first \( n \) terms of an A.P. is: \[ A_n = S_n = \frac{n}{2}(2a + (n - 1)d) \] Substituting \( a = 3 \), \( d = 4 \): \[ A_n = \frac{n}{2}(6 + 4(n - 1)) = \frac{n}{2}(6 + 4n - 4) = \frac{n}{2}(4n + 2) = n(2n + 1) \]

 Step 2: Target Expression

Find: \[ \frac{1}{25} \sum_{n=1}^{25} A_n = \frac{1}{25} \sum_{n=1}^{25} n(2n + 1) \] Break it as: \[ \frac{1}{25} \left( 2 \sum_{n=1}^{25} n^2 + \sum_{n=1}^{25} n \right) \]

 Step 3: Use Summation Formulas

\[ \sum_{n=1}^{25} n^2 = \frac{25 \cdot 26 \cdot 51}{6} = 5525 \quad \Rightarrow \quad 2 \sum n^2 = 11050 \] \[ \sum_{n=1}^{25} n = \frac{25 \cdot 26}{2} = 325 \] So: \[ \sum n(2n + 1) = 11050 + 325 = 11375 \] Final step: \[ \frac{1}{25} \cdot 11375 = \boxed{455} \]

 Final Answer

\[ \boxed{455} \]