Question
In an AP, S₅ = 75 and S₇ = 119, where S_n is sum of
first n terms. Find the 10th term.Solution
ATQ,
S_n = n/2[2a + (n−1)d]. S₅ = 5/2(2a + 4d) = 75 ⇒ 2a + 4d = 30. S₇ = 7/2(2a + 6d) = 119 ⇒ 2a + 6d = 34. Subtract: 2d = 4 ⇒ d = 2. Then 2a + 8 = 30 ⇒ a = 11. 10th term = a + 9d = 11 + 18 = 29.
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