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

Answer: (C)

455

Answer 2

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.