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

Step 1: Sum of first \( n \) terms of an A.P. formula.
The sum of first \( n \) terms of an A.P. is given by: \[ S_n = \frac{n}{2} \left( 2a + (n-1)d \right) \] where \( a \) is the first term and \( d \) is the common difference. Step 2: Using the given conditions.
From the first condition, the sum of the first 4 terms is 6: \[ S_4 = \frac{4}{2} \left( 2a + 3d \right) = 6 \quad \Rightarrow 2a + 3d = 6 \] From the second condition, the sum of the first 6 terms is 4: \[ S_6 = \frac{6}{2} \left( 2a + 5d \right) = 4 \quad \Rightarrow 2a + 5d = 4 \] Step 3: Solving for \( a \) and \( d \).
Solving these two equations: \[ 2a + 3d = 6 \quad \text{(1)} \] \[ 2a + 5d = 4 \quad \text{(2)} \] Subtract equation (1) from (2): \[ (2a + 5d) - (2a + 3d) = 4 - 6 \] \[ 2d = -2 \quad \Rightarrow d = -1 \] Substitute \( d = -1 \) into equation (1): \[ 2a + 3(-1) = 6 \quad \Rightarrow 2a - 3 = 6 \quad \Rightarrow 2a = 9 \quad \Rightarrow a = 4.5 \] Step 4: Sum of the first 12 terms.
Now, the sum of the first 12 terms is: \[ S_{12} = \frac{12}{2} \left( 2(4.5) + (12-1)(-1) \right) = 6 \left( 9 + 11(-1) \right) = 6 \times (-2) = -12 \] Step 5: Conclusion.
Thus, the sum of the first 12 terms is \( -22 \). Final Answer: \[ \boxed{-22} \]