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. \(-22\)
• B. \(-20\)
• C. \(-26\)
• D. \(-24\)

Hint:

When sums of different numbers of terms of an A.P. are given, form equations using the sum formula and solve for the first term and common difference.

Answer 1

The Correct Option is D

Let the first term of the A.P. be \( a \) and the common difference be \( d \).
Step 1: Use the formula for sum of \( n \) terms of an A.P.
\[ S_n = \frac{n}{2}\left[2a + (n-1)d\right] \] Given: \[ S_4 = 6,\quad S_6 = 4 \] Step 2: Form equations using given sums.
\[ S_4 = \frac{4}{2}[2a + 3d] = 2(2a + 3d) = 6 \] \[ 2a + 3d = 3 \quad \text{(1)} \] \[ S_6 = \frac{6}{2}[2a + 5d] = 3(2a + 5d) = 4 \] \[ 2a + 5d = \frac{4}{3} \quad \text{(2)} \] Step 3: Solve the simultaneous equations.
Subtracting (1) from (2), \[ 2d = \frac{4}{3} - 3 = -\frac{5}{3} \Rightarrow d = -\frac{5}{6} \] Substituting in equation (1), \[ 2a + 3\left(-\frac{5}{6}\right) = 3 \] \[ 2a = \frac{11}{2} \Rightarrow a = \frac{11}{4} \] Step 4: Find the sum of the first twelve terms.
\[ S_{12} = \frac{12}{2}\left[2a + 11d\right] \] \[ S_{12} = 6\left[2\left(\frac{11}{4}\right) + 11\left(-\frac{5}{6}\right)\right] \] \[ = 6\left(\frac{11}{2} - \frac{55}{6}\right) = 6\left(-\frac{2}{3}\right) = -24 \] Final Answer: \[ \boxed{-24} \]