In an Arithmetic Progression (AP), the $n$-th term $T_n$ is given by the formula:

$T_n = a + (n - 1) d$

where:

• $a$ is the first term,
• $d$ is the common difference.

Step 1: Translate the problem into equations.

From the problem:

• The 16th term is three times the 5th term:
  $T_{16} = 3T_5$ Using the formula for the $n$-th term, this becomes: $a + 15d = 3(a + 4d)$

• The 12th term is 100: $T_{12} = 100 \quad \Rightarrow \quad a + 11d = 100$

Step 2: Solve the system of equations.

Equation 1 (from $T_{16} = 3T_5$):

$a + 15d = 3(a + 4d)$ Expanding the right side: $a + 15d = 3a + 12d$ Simplify: $15d - 12d = 3a - a \quad \Rightarrow \quad 3d = 2a$ Thus, we have: a = \frac{3d}{2} \tag{1}

Equation 2 (from $T_{12} = 100$):

$a + 11d = 100$ Substitute $a = \frac{3d}{2}$ into this equation: $\frac{3d}{2} + 11d = 100$ Multiply through by 2 to eliminate the fraction: $3d + 22d = 200 \quad \Rightarrow \quad 25d = 200$ Solve for $d$: $d = 8$

Step 3: Find $a$.

Substitute $d = 8$ into equation (1): $a = \frac{3 \times 8}{2} = 12$

Step 4: Find the sum of the first 20 terms.

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

For $n = 20$, $a = 12$, and $d = 8$, we have: $S_{20} = \frac{20}{2} [2 \times 12 + (20 - 1) \times 8]$ Simplify: [ S_{20} = 10 [24 + 19 \times 8] = 10 [24 + 152] = 10 \times 176 = 1760 ]

Final Answer:

The sum of the first 20 terms is $1760$.

Would you like further details or have any other questions?

Here are five related questions to expand your understanding:

1. How do you find the common difference in an AP if only the first and last terms are known?
2. What is the formula for the sum of an AP when you know the first and last terms?
3. How can you derive the $n$-th term of a geometric progression?
4. What is the sum of the first 50 terms of the AP in this problem?
5. How would the sum formula change if the series was in geometric progression?

Tip: Always double-check which term position is given in the problem to avoid mixing up indices (e.g., 12th term versus 12 terms).