Question

If the sum of the first four terms of an A.P. is 6 and the sum of its first six terms is 4 , then the sum of its first twelve terms is

• A. -20
• B. -24
• C. -26
• D. -22

Hint:

When working with A.P. sums, you can simplify equations before solving. Notice how \(S_n = \dots\) directly translates to a linear equation in \(a\) and \(d\).

Answer 1

The Correct Option is D

To solve this problem, we need to use the formula for the sum of the first \(n\) terms of an arithmetic progression (A.P.). The sum \(S_n\) of the first \(n\) terms of an A.P. is given by:

\(S_n = \frac{n}{2} \times (2a + (n-1)d)\)

where \(a\) is the first term and \(d\) is the common difference of the A.P.

1. According to the problem, the sum of the first four terms \((S_4)\) is 6. Therefore:

\(\frac{4}{2} \times (2a + 3d) = 6\)

Simplifying:

\(2 \times (2a + 3d) = 6\)

\(2a + 3d = 3 \quad \text{(Equation 1)}\)

1. The sum of the first six terms \((S_6)\) is 4. Therefore:

\(\frac{6}{2} \times (2a + 5d) = 4\)

Simplifying:

\(3 \times (2a + 5d) = 4\)

\(2a + 5d = \frac{4}{3} \quad \text{(Equation 2)}\)

1. Now, subtract Equation 1 from Equation 2 to find the common difference \(d\):

\((2a + 5d) - (2a + 3d) = \frac{4}{3} - 3\)

\(2d = \frac{4}{3} - \frac{9}{3} = -\frac{5}{3}\)

\(d = -\frac{5}{6}\)

1. Substitute \(d = -\frac{5}{6}\) back into Equation 1 to find \(a\):

\(2a + 3 \times -\frac{5}{6} = 3\)

\(2a - \frac{15}{6} = 3\)

\(2a = 3 + \frac{15}{6} = \frac{18}{6} + \frac{15}{6} = \frac{33}{6}\)

\(a = \frac{33}{12} = \frac{11}{4}\)

1. Finally, find the sum of the first twelve terms \((S_{12})\):

\(S_{12} = \frac{12}{2} \times (2a + 11d)\)

\(= 6 \times (2 \times \frac{11}{4} + 11 \times -\frac{5}{6})\)

\(= 6 \times (\frac{22}{4} - \frac{55}{6})\)

\(= 6 \times (\frac{33}{6} - \frac{55}{6})\)

\(= 6 \times -\frac{22}{6} = -22\)

Therefore, the sum of the first twelve terms is -22. Hence, the correct answer is:

-22