Question

Let \( x_1, x_2, \dots, x_n \) be in an A.P. If \( x_1 + x_4 + x_9 + x_{11} + x_{20} + x_{22} + x_{27} + x_{30} = 272 \), then \( x_1 + x_2 + x_3 + \dots + x_{30} \) is equal to:

• A. \( 1020 \)
• B. \( 1200 \)
• C. \( 716 \)
• D. \( 2720 \)
• E. \( 2072 \)

Hint:

Always check if the indices of the given terms add up to \( n + 1 \). Here, \( 1+30=31 \), \( 4+27=31 \), \( 9+22=31 \), and \( 11+20=31 \). This confirms you can use the equidistant property.

Answer 1

The Correct Option is A

Concept: In an Arithmetic Progression (A.P.) of \( n \) terms, the sum of terms equidistant from the beginning and the end is constant: \[ x_1 + x_n = x_2 + x_{n-1} = x_3 + x_{n-2} = \dots \]

Step 1: Pair the terms from the given equation.
The series has 30 terms. We can pair terms \( x_k \) and \( x_{30-k+1} \):
• \( x_1 + x_{30} = S_{pair} \)
• \( x_4 + x_{27} = S_{pair} \)
• \( x_9 + x_{22} = S_{pair} \)
• \( x_{11} + x_{20} = S_{pair} \)

Step 2: Solve for the sum of a single pair.
The given sum of 8 terms consists of 4 such pairs: \[ (x_1 + x_{30}) + (x_4 + x_{27}) + (x_9 + x_{22}) + (x_{11} + x_{20}) = 272 \] \[ 4(x_1 + x_{30}) = 272 \quad \Rightarrow \quad x_1 + x_{30} = 68 \]

Step 3: Calculate the sum of the first 30 terms (\( S_{30} \)).
The formula for the sum of \( n \) terms is \( S_n = \frac{n}{2}(x_1 + x_n) \). \[ S_{30} = \frac{30}{2}(x_1 + x_{30}) = 15 \times 68 \] \[ S_{30} = 1020 \]