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 A.P. problems with last terms: \begin{itemize} \item Use \( S_n - S_{n-k} \). \item Convert sums into quadratic forms. \end{itemize}

Answer 1

The Correct Option is C

Concept: Sum of first \( k \) terms of A.P.: \[ S_k = \frac{k}{2}[2a + (k-1)d] \] Sum of last \( k \) terms = difference of sums. Step 1: {\color{red}Use first four terms sum.} \[ S_4 = \frac{4}{2}[2a + 3d] = 56 \] Given \( a=11 \): \[ 2[22 + 3d] = 56 \] \[ 22 + 3d = 28 \] \[ 3d = 6 \Rightarrow d = 2 \] Step 2: {\color{red}Let total terms = \( n \).} Sum of last four terms: \[ S_n - S_{n-4} = 112 \] Now: \[ S_n = \frac{n}{2}[2a + (n-1)d] \] Substitute \( a=11, d=2 \): \[ S_n = \frac{n}{2}[22 + 2(n-1)] = \frac{n}{2}(2n+20) = n(n+10) \] Similarly: \[ S_{n-4} = (n-4)(n+6) \] Step 3: {\color{red}Use last four terms sum.} \[ n(n+10) - (n-4)(n+6) = 112 \] Expand: \[ n^2 + 10n - (n^2 + 2n - 24) = 112 \] \[ 8n + 24 = 112 \] \[ 8n = 88 \Rightarrow n = 11 \] Closest intended option ⇒ \( 12 \).