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

The given sequence is: 3, 7, 11, 15, ...

This is an arithmetic progression (AP) with first term (a) = 3 and common difference (d) = 7 - 3 = 4.

---

For any arithmetic progression, the sum of the first n terms is:

$A_n = \dfrac{n}{2}[2a + (n-1)d]$

Substituting our values (a = 3, d = 4):

$A_n = \dfrac{n}{2}[2 \times 3 + (n-1) \times 4]$

$A_n = \dfrac{n}{2}[6 + 4n - 4]$

$A_n = \dfrac{n}{2}[2 + 4n]$

$A_n = \dfrac{n}{2} \times 2(1 + 2n)$

$A_n = n(1 + 2n) = n + 2n^2$

Therefore: $A_n = 2n^2 + n$

---

We need to find: $\dfrac{1}{25} \sum_{n=1}^{25} A_n$

$\dfrac{1}{25} \sum_{n=1}^{25} A_n = \dfrac{1}{25} \sum_{n=1}^{25} (2n^2 + n)$

$= \dfrac{1}{25} \left[2\sum_{n=1}^{25} n^2 + \sum_{n=1}^{25} n\right]$

---

Using standard summation formulas:

$\sum_{n=1}^{k} n = \dfrac{k(k+1)}{2}$ (sum of first k natural numbers)

$\sum_{n=1}^{k} n^2 = \dfrac{k(k+1)(2k+1)}{6}$ (sum of squares of first k natural numbers)

For k = 25:

$\sum_{n=1}^{25} n = \dfrac{25 \times 26}{2} = 325$

$\sum_{n=1}^{25} n^2 = \dfrac{25 \times 26 \times 51}{6} = \dfrac{33150}{6} = 5525$

---

$\dfrac{1}{25} \left[2 \times 5525 + 325\right]$

$= \dfrac{1}{25} \left[11050 + 325\right]$

$= \dfrac{1}{25} \times 11375$

$= 455$

Answer: 455

This problem combines arithmetic progression sums with summation formulas. The trick is recognizing that we need to find $A_n$ first, then use standard summation formulas to evaluate the final expression efficiently.