How many terms of the ap 63 60 57
Asked by admin @ in Math viewed by 292 People
How many terms of AP 63,60 ,57 must be so that their sum is 693?
Answered by admin @
Given AP is 63,60,57...
a = 63, d = 60 - 63 = -3.
We know that sn = n/2(2a + (n-1)d)
693 = n/2(2(63) + (n-1)*(-3))
693 = n/2(126 - 3n + 3)
1386 = n(129-3n)
3n^2 - 129n + 1386 = 0
3n^2 - 66n - 63n + 1386 = 0
3n(n - 22) - 63(n-22) = 0
(3n - 63) = 0 (or) (n-22) = 0
n = 21 (or) 22.
Hope this helps! :-))
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