how many terms of AP 54,51,48,45,... be taken so that their sum is 513?
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Here, a = 54 and d = -3
we know that,
S = n/2 (2a + (n-1)d)
so, 513 = n/2 (108 + (n-1)-3)
=> 1026 = n (108 - 3n +3)
=> 1026 = n (111 - 3n)
=> 1026 = -3n^2 + 111n
=> 0 = 3n^2 - 111n + 1026
=> 0 = n^2 - 37n + 342
By mid term splitting,
n = 18 or n = 19
we know that,
S = n/2 (2a + (n-1)d)
so, 513 = n/2 (108 + (n-1)-3)
=> 1026 = n (108 - 3n +3)
=> 1026 = n (111 - 3n)
=> 1026 = -3n^2 + 111n
=> 0 = 3n^2 - 111n + 1026
=> 0 = n^2 - 37n + 342
By mid term splitting,
n = 18 or n = 19
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