Understanding the Problem

We are given an arithmetic progression (AP) with 100 terms, where the first term a₁ = 3. We define S_p as the sum of the first p terms of this AP, for 1 ≤ p ≤ 100. For any integer n between 1 and 20, we let m = 5n. We are told that the ratio S_m/S_n does not depend on the value of n. This condition must be used to find the common difference d, which will allow us to calculate the second term a₂ = a₁ + d = 3 + d.

Step-by-Step Solution

Step 1: General Formulas for an Arithmetic Progression

For any arithmetic progression, the n^th term is given by:

$a_n=a_1+(n-1)d$
where $d$ is the common difference.

The sum of the first n terms is given by:

$S_n=\frac{n}{2}[2a_1+(n-1\left)d\right]$

In our case, a₁ = 3, so this simplifies to:

$S_n=\frac{n}{2}[6+(n-1\left)d\right]$

Step 2: Write the Expression for the Ratio S_m/S_n

We are given that m = 5n. Therefore, we can write the sum S_m as:

$S_m=S_{5n}=\frac{5n}{2}[6+(5n-1\left)d\right]$

Now, let's form the required ratio:

$\frac{S_m}{S_n}=\frac{\frac{5n}{2}[6+(5n-1\left)d\right]}{\frac{n}{2}[6+(n-1\left)d\right]}$

The factors of n/2 in the numerator and denominator cancel out, simplifying the expression to:

$\frac{S_m}{S_n}=5\cdot \frac{6+(5n-1)d}{6+(n-1)d}$
Let's denote this ratio as R.

Step 3: Apply the Condition That the Ratio is Independent of n

For R to be independent of n, the variable 'n' must cancel out from the expression. This means the expression must be of the form constant × (constant). The only way for the 'n' terms to cancel is if the coefficients of 'n' in the numerator and denominator are proportional.

Let's write the numerator and denominator, focusing on the terms that contain 'n':

Numerator: $6+5d\cdot n-d$     Denominator: $6+d\cdot n-d$

For the ratio to be independent of n, the ratio of the coefficient of n in the numerator to the coefficient of n in the denominator must be equal to the ratio of the constant terms. This is a standard condition for such expressions to be independent of the variable.

Therefore, we set up the following equation:

$\frac{5d}{d}=\frac{6-d}{6-d}$

The right side simplifies to 1. The left side simplifies to 5. This gives us 5 = 1, which is a contradiction.

This contradiction implies that our initial approach needs a refinement. The condition for independence from n is actually that the denominator must be a factor of the numerator. In other words, the numerator must be exactly 5 times the denominator for the 'n' to cancel perfectly.

So, we require: $6+(5n-1)d=5\cdot [6+(n-1\left)d\right]$ for the ratio to be a constant (which would be 5 × 5 = 25).

Step 4: Solve the Equation for the Common Difference d

Let's solve the equation from the refined condition:

$6+(5n-1)d=30+5(n-1)d$
$6+5nd-d=30+5nd-5d$

The term $5$