Question:medium
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:
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:
Show 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.
Updated On: May 6, 2026
- \( 1020 \)
- \( 1200 \)
- \( 716 \)
- \( 2720 \)
- \( 2072 \)
Show Solution
The Correct Option is A
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