Question:
Let \(a_1, a_2, \ldots, a_n\) be a given A.P. whose common difference is an integer and \(S_n = a_1 + a_2 + \ldots + a_n\). If \(a_1 = 1\), \(a_n = 300\) and \(15 \le n \le 50\), then the ordered pair \((S_{n-4},\ a_{n-4})\) is equal to:
Let \(a_1, a_2, \ldots, a_n\) be a given A.P. whose common difference is an integer and \(S_n = a_1 + a_2 + \ldots + a_n\). If \(a_1 = 1\), \(a_n = 300\) and \(15 \le n \le 50\), then the ordered pair \((S_{n-4},\ a_{n-4})\) is equal to:
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\(d = \frac{299}{n-1}\) must be an integer and \(299 = 13 \times 23\). Use the range of n to fix n and d.
Updated On: Nov 5, 2026
- (2480, 249)
- (2480, 248)
- (2490, 248)
- (2490, 249)
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The Correct Option is C
Solution and Explanation
\(a_n = 1 + (n-1)d = 300 \Rightarrow (n-1)d = 299 = 13\times 23\).
d is an integer, so \(n-1\) is a divisor of 299: 1, 13, 23, 299.
With \(15 \le n \le 50\), i.e. \(14 \le n-1 \le 49\): \(n-1 = 23\), so \(n = 24\) and \(d = 13\).
\(n-4 = 20\): \(a_{20} = 1 + 19\times 13 = 248\).
\(S_{20} = \frac{20}{2}(a_1 + a_{20}) = 10(1+248) = 2490\).
The pair is (2490, 248).
Answer: (3) (2490, 248)
d is an integer, so \(n-1\) is a divisor of 299: 1, 13, 23, 299.
With \(15 \le n \le 50\), i.e. \(14 \le n-1 \le 49\): \(n-1 = 23\), so \(n = 24\) and \(d = 13\).
\(n-4 = 20\): \(a_{20} = 1 + 19\times 13 = 248\).
\(S_{20} = \frac{20}{2}(a_1 + a_{20}) = 10(1+248) = 2490\).
The pair is (2490, 248).
Answer: (3) (2490, 248)
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