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

The correct answer is C:455
Given the arithmetic progression: 3,7,11,..., where \(A_n\) denotes the sum of the first n terms of this progression. 
Common difference (d) between consecutive terms:7-3=4. 
The nth term \((a_n)\) of an arithmetic progression:\(a_n=a_1+(n-1)\times{d}\). 
Substitute values (\(a_1=3\), d = 4, n is term number): 
\(a_n=3+(n-1)\times4\), 
\(a_n=4n-1\). 
Sum of the first n terms \((A_n)\) of an arithmetic progression: 
\(A_n=\frac{n}{2}\times[2a_1+(n-1)d]\). 
Substitute values and simplify: 
\(A_n=\frac{n}{2}\times(2\times3+(n-1)\times4)\), 
\(A_n=\frac{n}{2}\times(6+4n-4)\), 
\(A_n=\frac{n}{2}\times(4n+2)\), 
\(A_n = 2n^2 + n\). 
Calculate the value of \(\frac{1}{25}∑^{25}_{n=1} A_n\): 
\((\frac{1}{25})Σ^{25}_{n=1}(A_n)=(\frac{1}{25})\times(A_1 + A_2 + A_3 + ... + A_25)\). 
Substitute expression for An into the sum: 
\((\frac{1}{25}) \times (2 \times 1^2 + 1 + 2 \times 2^2 + 2 + 2\times3^2 + 3 + ... + 2\times25^2+25)\). 
Simplify using sum of squares and sum of natural numbers formulas: 
\((\frac{1}{25}) \times (2 \times \frac{(25 * (25 + 1) * (2 * 25 + 1))}{6}+\frac{(25\times (25 + 1))}{2})\). 
Further simplification: 
\(2 \times \frac{(25 \times 26 \times 51)}{6}+\frac{(25 \times 26)}{2}\), 
\(2 \times 25 \times 13 \times 17 + 25 \times 13\), 
\(25 \times 13 \times (2 \times 17 + 1)\), 
\(25 \times 13 \times 35\). 
Finally,the value of \((\frac{1}{25}) \times Σ^{25}_{n=1}(A_n)\) is \(13 \times 35 = 455\). 
So, the answer is c. 455. 

Answer 2

Total number of terms in an A.P.
\(A_n = \frac{n}{2}(2a + (n - 1)d)\)

\(\frac{1}{25} \sum_{n=1}^{25} A_n\)

=\(\frac{1}{25} \sum_{n=1}^{25} \frac{n}{2}(2a+(n-1)d)\)

=\(\frac{1}{25} \sum_{n=1}^{25} \frac{n}{2}(6+(n-1)4)\)

=\(\frac{1}{25} \sum_{n=1}^{25} n(2n+1)\)

=\(\frac{1}{25}\left(2\sum_{n=1}^{25}n^2 + \sum_{n=1}^{25}n\right)\)

=\(\frac{1}{25}\left(2 \times \frac{25 \times 26 \times 51}{6} + \frac{25 \times 26}{2}\right) \)
\(=2×17×13+13=2×17×13+13\)
\(=35×13=35×13\)
\(= 455\)