Step-by-Step Solution
Key Concept: Use the formula for the sum of first n terms of an AP: \(S_n = \frac{n}{2}[2a+(n-1)d]\). Set up two equations using the given sums (for n = 3 and n = 6) and solve the simultaneous linear equations to obtain the first term \(a\) and common difference \(d\). The required three terms are then \(a, a+d, a+2d\).
1. Let the first term be \(a\) and the common difference be \(d\).\
2. Sum of first three terms (\(S_3\)) is given as 12. Using the sum formula:
$$S_3 = \frac{3}{2}[2a+(3-1)d] = \frac{3}{2}(2a+2d) = 3(a+d) = 12$$
Hence, \(a + d = 4\).\
3. Sum of first six terms (\(S_6\)) is given as 42. Using the sum formula:
$$S_6 = \frac{6}{2}[2a+(6-1)d] = 3(2a+5d) = 42$$
Hence, \(2a + 5d = 14\).\
4. Solve the two linear equations:
\[\begin{cases} a + d = 4 \ 2a + 5d = 14 \end{cases}\]
Multiply the first equation by 2: \(2a + 2d = 8\).
Subtract from the second equation:
\( (2a+5d) - (2a+2d) = 14 - 8 \)\
\(3d = 6 \Rightarrow d = 2\).
Substitute \(d = 2\) into \(a + d = 4\): \(a + 2 = 4 \Rightarrow a = 2\).
5. The first three terms of the AP are:
\[a = 2,\quad a+d = 2+2 = 4,\quad a+2d = 2+2\times2 = 6\].
Correct Answer: The first three terms are 2, 4, 6.