Question

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

• A. \(-22\)
• B. \(-21\)
• C. \(-23\)
• D. \(-24\)

Hint:

When two different sums of an A.P. are given, form equations using the sum formula and solve them simultaneously to find \(a\) and \(d\).

Answer 1

The Correct Option is A

Concept: For an arithmetic progression with first term \(a\) and common difference \(d\), the sum of first \(n\) terms is given by: \[ S_n = \frac{n}{2}\bigl(2a + (n-1)d\bigr) \]
Step 1: Use the given condition \(S_4 = 6\). \[ S_4 = \frac{4}{2}(2a + 3d) = 6 \] \[ 2(2a + 3d) = 6 \Rightarrow 2a + 3d = 3 \quad \cdots (1) \]
Step 2: Use the given condition \(S_6 = 4\). \[ S_6 = \frac{6}{2}(2a + 5d) = 4 \] \[ 3(2a + 5d) = 4 \Rightarrow 2a + 5d = \frac{4}{3} \quad \cdots (2) \]
Step 3: Subtract equation (2) from equation (1). \[ (2a + 3d) - (2a + 5d) = 3 - \frac{4}{3} \] \[ -2d = \frac{5}{3} \Rightarrow d = -\frac{5}{6} \]
Step 4: Find the first term \(a\). Substitute \(d = -\frac{5}{6}\) in equation (1): \[ 2a + 3\left(-\frac{5}{6}\right) = 3 \] \[ 2a - \frac{15}{6} = 3 \] \[ 2a = \frac{33}{6} = \frac{11}{2} \Rightarrow a = \frac{11}{4} \]
Step 5: Find the sum of first 12 terms. \[ S_{12} = \frac{12}{2}(2a + 11d) \] \[ S_{12} = 6\left(\frac{11}{2} + 11\left(-\frac{5}{6}\right)\right) \] \[ S_{12} = 6\left(\frac{33 - 55}{6}\right) = 6\left(-\frac{22}{6}\right) = -22 \]