Step 1: Understanding the Question:
We are given the sum of the first 4 terms and the first 6 terms of an Arithmetic Progression (A.P.). We need to find the first term (\(a\)) and the common difference (\(d\)) to calculate the sum of the first 12 terms.
Step 2: Key Formula or Approach:
The sum of the first \(n\) terms of an A.P. is given by:
\[ S_n = \frac{n}{2} [2a + (n-1)d] \]
We will set up two linear equations using the given data and solve for \(a\) and \(d\).
Step 3: Detailed Explanation:
Given \(S_4 = 6\):
\[ \frac{4}{2} [2a + (4-1)d] = 6 \]
\[ 2 [2a + 3d] = 6 \]
\[ 2a + 3d = 3 \quad \dots \text{(Equation 1)} \]
Given \(S_6 = 4\):
\[ \frac{6}{2} [2a + (6-1)d] = 4 \]
\[ 3 [2a + 5d] = 4 \]
\[ 6a + 15d = 4 \quad \dots \text{(Equation 2)} \]
To solve these equations, multiply Equation 1 by 3:
\[ 6a + 9d = 9 \quad \dots \text{(Equation 3)} \]
Subtract Equation 3 from Equation 2:
\[ (6a + 15d) - (6a + 9d) = 4 - 9 \]
\[ 6d = -5 \implies d = -\frac{5}{6} \]
Substitute the value of \(d\) into Equation 1:
\[ 2a + 3\left(-\frac{5}{6}\right) = 3 \]
\[ 2a - \frac{5}{2} = 3 \]
\[ 2a = 3 + \frac{5}{2} = \frac{11}{2} \implies a = \frac{11}{4} \]
Now, we find the sum of the first 12 terms (\(S_{12}\)):
\[ S_{12} = \frac{12}{2} [2a + (12-1)d] \]
\[ S_{12} = 6 [2a + 11d] \]
Substitute \(a = 11/4\) and \(d = -5/6\):
\[ S_{12} = 6 \left[ 2\left(\frac{11}{4}\right) + 11\left(-\frac{5}{6}\right) \right] \]
\[ S_{12} = 6 \left[ \frac{11}{2} - \frac{55}{6} \right] \]
Find a common denominator for the terms inside the bracket:
\[ S_{12} = 6 \left[ \frac{33 - 55}{6} \right] \]
\[ S_{12} = 33 - 55 = -22 \]
Step 4: Final Answer:
The sum of the first twelve terms is -22.