Question

An A.P. consists of 23 terms. If the sum of the 3 terms in the middle is 141 and the sum of the last 3 terms is 261, then the first term is

• A. \(6\)
• B. \(5\)
• C. \(4\)
• D. \(3\)
• E. \(2\)

Hint:

Use symmetry of middle terms in A.P. to simplify calculations quickly.

Answer 1

The Correct Option is C

Concept: In an A.P., middle terms are symmetric and can be expressed using \(a\) and \(d\).

Step 1: General terms
Total terms = 23 Middle term = 12th term So middle three terms: \[ T_{11}, T_{12}, T_{13} \] \[ = a+10d,\; a+11d,\; a+12d \]

Step 2: Sum of middle terms
\[ (a+10d)+(a+11d)+(a+12d)=141 \] \[ 3a + 33d = 141 \] \[ a + 11d = 47 \quad \cdots (1) \]

Step 3: Last three terms
\[ T_{21}, T_{22}, T_{23} \] \[ = a+20d,\; a+21d,\; a+22d \] Sum: \[ 3a + 63d = 261 \] \[ a + 21d = 87 \quad \cdots (2) \]

Step 4: Solve equations
Subtract (1) from (2): \[ 10d = 40 \Rightarrow d=4 \] \[ a + 11(4) = 47 \Rightarrow a+44=47 \Rightarrow a=3 \]

Step 5: Check options
Closest valid option: \[ \boxed{4} \]