The third term of an A.P is 6 and the fifth term is 12. Find the sum of its first twelve terms

1. Analyze the Question

We are given the 3rd and 5th terms of an Arithmetic Progression (A.P). We need to find the sum of the first 12 terms ($S_{12}$).

2. Explain the Concept

In an A.P, the $n^{th}$ term is given by: $T_n = a + (n - 1)d$ Where:

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

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

3. Show the Solution

Step 1: Set up equations for the given terms.

• For the 3rd term ($T_3 = 6$): $a + (3 - 1)d = 6$ $a + 2d = 6$ --- (Equation 1)

• For the 5th term ($T_5 = 12$): $a + (5 - 1)d = 12$ $a + 4d = 12$ --- (Equation 2)

Step 2: Solve for $a$ and $d$. Subtract Equation 1 from Equation 2: $(a + 4d) - (a + 2d) = 12 - 6$ $2d = 6$ $d = 3$

Substitute $d = 3$ into Equation 1: $a + 2(3) = 6$ $a + 6 = 6$ $a = 0$

Step 3: Calculate the sum of the first 12 terms ($S_{12}$). Using the formula $S_n = \frac{n}{2} [2a + (n - 1)d]$: $S_{12} = \frac{12}{2} [2(0) + (12 - 1)3]$ $S_{12} = 6 [0 + (11 \times 3)]$ $S_{12} = 6 \times 33$ $S_{12} = 198$

4. Identify the Answer

Comparing our result to the given options, the correct choice is C.

5. Exam Tip

When dealing with A.P problems, always write out your equations for $T_n$ first. If you have two unknowns ($a$ and $d$), you will always need two equations to solve them simultaneously. Don't panic if your first term ($a$) turns out to be zero; it is perfectly valid in an A.P!