The sum of the first 24 terms of an AP is \(\frac{12504}{25}\) and the sum of the next 24 terms is \(\frac{17112}{25}\). What is the \(3^{\text{rd}}\) term?
Show Hint
- \(\frac{8}{25}\)
- \(\frac{429}{25}\)
- \(\frac{89}{5}\)
- \(\frac{441}{25}\)
The Correct Option is C
Solution and Explanation
Let \(S_{24}\) be sum of first 24 terms. \(S_{48}\) be sum of first 48 terms. \(S_{24} = \frac{12504}{25}\). Sum of next 24 terms = \(S_{48} - S_{24} = \frac{17112}{25}\). So, \(S_{48} = \frac{12504 + 17112}{25} = \frac{29616}{25}\). Step 2: Set up Equations:
Formula: \(S_n = \frac{n}{2}[2a + (n-1)d]\). 1) \(12(2a + 23d) = \frac{12504}{25} \implies 2a + 23d = \frac{1042}{25}\) 2) \(24(2a + 47d) = \frac{29616}{25} \implies 2a + 47d = \frac{1234}{25}\) Step 3: Solve for d and a:
Subtract (1) from (2): \(24d = \frac{1234 - 1042}{25} = \frac{192}{25}\). \(d = \frac{192}{25 \times 24} = \frac{8}{25}\). Substitute \(d\) into (1): \(2a + 23(\frac{8}{25}) = \frac{1042}{25}\) \(2a = \frac{1042 - 184}{25} = \frac{858}{25} \implies a = \frac{429}{25}\). Step 4: Find 3rd Term:
\(T_3 = a + 2d = \frac{429}{25} + \frac{16}{25} = \frac{445}{25}\). Simplify: \(\frac{445}{25} = \frac{89}{5}\).
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