Question

If sum of first 4 terms of an A.P. is 6 and sum of first 6 terms is 4, then sum of first 12 terms of an A.P. is

• A. -22
• B. -21
• C. -23
• D. -24

Hint:

To solve for the sum of an arithmetic progression, use the formula for the sum of the first \( n \) terms and solve for the unknowns.

Answer 1

The Correct Option is A

To find the sum of the first 12 terms of the given Arithmetic Progression (A.P.), let's first understand the given information and use it systematically.

An A.P. is defined by its first term \(a\) and common difference \(d\). The sum of the first \(n\) terms of an A.P. is given by the formula:

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

Let's use the information provided in the problem statement:

• The sum of the first 4 terms is 6: \(S_4 = 6\)
• The sum of the first 6 terms is 4: \(S_6 = 4\)

Using the sum formula for the first 4 terms:

\(S_4 = \frac{4}{2} \times (2a + 3d) = 6\)

This simplifies to:

\(2 \times (2a + 3d) = 6 \implies 2a + 3d = 3 \quad \text{(Equation 1)}\)

Using the sum formula for the first 6 terms:

\(S_6 = \frac{6}{2} \times (2a + 5d) = 4\)

This simplifies to:

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

We now have two equations:

1. \(2a + 3d = 3\)
2. \(2a + 5d = \frac{4}{3}\)

Subtract Equation 1 from Equation 2 to eliminate \(2a\):

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

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

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

Substitute the value of \(d\) back into Equation 1:

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

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

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

\(2a = \frac{11}{2} \implies a = \frac{11}{4}\)

Finally, to find the sum of the first 12 terms:

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

\(S_{12} = 6 \times (2 \times \frac{11}{4} + 11 \times -\frac{5}{6})\)

\(S_{12} = 6 \times (\frac{22}{4} - \frac{55}{6})\)

Finding a common denominator (12) yields:

\(\frac{22}{4} = \frac{66}{12}\) and \(\frac{55}{6} = \frac{110}{12}\)

\(S_{12} = 6 \times \left(\frac{66}{12} - \frac{110}{12}\right) = 6 \times \left(-\frac{44}{12}\right) = 6 \times \left(-\frac{11}{3}\right)\)

\(S_{12} = -22\)

Therefore, the sum of the first 12 terms of the A.P. is \(-22\).