Question

The sum of the first four terms of an arithmetic progression is 56. The sum of the last four terms is 112. If its first term is 11, then the number of terms is:

• A. 10
• B. 11
• C. 12
• D. 13

Hint:

For problems involving the sum of specific terms in an arithmetic progression, use the general sum formula for arithmetic progressions: \[ S_n = \frac{n}{2} \times (2a + (n-1)d) \] Also, remember that the sum of consecutive terms is another way to work with the properties of an arithmetic progression.

Answer 1

The Correct Option is B

Let the first term of the arithmetic progression be \( a = 11 \), and the common difference be \( d \).

Step 1: Calculate the common difference.
The sum of the first four terms, \( S_4 \), is 56. \[\nS_4 = 4 \times \frac{2a + 3d}{2} = 56\n\] Substitute \( a = 11 \): \[\n4 \times \frac{2(11) + 3d}{2} = 56\n\] Simplifying: \[\n4 \times \frac{22 + 3d}{2} = 56\n\] \[\n2 \times (22 + 3d) = 56\n\] \[\n44 + 6d = 56\n\] \[\n6d = 12 \quad \Rightarrow \quad d = 2\n\]
Step 2: Determine the number of terms.
The sum of the last four terms is 112. The last four terms are: \[\na + (n-4)d, a + (n-3)d, a + (n-2)d, a + (n-1)d\n\] The sum of the last four terms is: \[\nS_{\text{last 4}} = 4 \times \frac{2(a + (n-4)d) + 3d}{2} = 112\n\] Substitute \( a = 11 \) and \( d = 2 \): \[\n4 \times \frac{2(11 + (n-4)2) + 3(2)}{2} = 112\n\] Simplifying: \[\n4 \times \frac{2(11 + 2n - 8) + 6}{2} = 112\n\] \[\n4 \times \frac{2(2n + 3) + 6}{2} = 112\n\] \[\n4 \times \frac{4n + 6 + 6}{2} = 112\n\] \[\n4 \times \frac{4n + 12}{2} = 112\n\] \[\n2 \times (4n + 12) = 112\n\] \[\n8n + 24 = 112\n\] \[\n8n = 88 \quad \Rightarrow \quad n = 11\n\]
Step 3: Conclusion.
Thus, the number of terms is \( \boxed{11} \).