The problem is written in Sinhala, so I will translate and solve it step-by-step. Here is the transcription and translation:

Problem Statement:

1. The sequence $T_n$ is an arithmetic progression (AP).
2. The sum of the first four terms of the AP is 12. $S_4 = 12$
3. The sum of the squares of the first two terms equals the square of the third term. $3T_3 = T_{11}$

Let's solve it step by step Let's analyze and solve the problem systematically.

Given:

1. $S_4 = 12$, the sum of the first four terms of an arithmetic progression.
2. $3T_3 = T_{11}$, where $T_n$ represents the $n$-th term of the arithmetic progression.

Step 1: Formula for the $n$-th term of an AP

The $n$-th term of an AP is given by: $T_n = a + (n-1)d$ Where:

• $a$ = the first term
• $d$ = common difference

Step 2: Formula for the sum of the first $n$ terms of an AP

The sum of the first $n$ terms is: $S_n = \frac{n}{2} [2a + (n-1)d]$

Step 3: Apply $S_4 = 12$

For $S_4$: $S_4 = \frac{4}{2} [2a + 3d] = 12$ Simplify: $2[2a + 3d] = 12$ [ 2