an arithmetic progression, the sum of first 15 terms equal to that of the first 35 tems. Find the sum of the first 50 terms.
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Given : an arithmetic progression, the sum of first 15 terms equal to that of the first 35 terms.
To Find : sum of the first 50 terms.
Solution:
a , a+ d , a + 2d
Sₙ = (n/2)(2a + (n - 1)d)
S₁₅ = (15/2)(2a + 14d) = 15(a + 7d)
S₃₅ = (35/2)(2a + 34d) = 35(a + 17d)
15(a + 7d) = 35(a + 17d)
=> 3(a + 7d) = 7(a + 17d)
=> 3a + 21d = 7a + 119d
=> 98d + 4a = 0
=> 49d+ 2a = 0
=> 2a + 49d = 0
S₅₀ = (50/2)(2a + 49d) = 25(0) = 0
sum of the first 50 terms. = 0
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