The number of terms needed to get Sn = 0 in the A.P of 63, 60, 57. . . . .
Here the sequence is 63, 60, 57.. . . . So a = 63, d = – 3 and n = ?, Sn= 0; Sn = (n/2) [2 x 63 + (n – 1)(-3)] = 0 ⇒ [126 – 3n + 3] = 0 ⇒ [129 – 3n] = 0 ⇒ n = 43 So in the given sequence 43 number of terms are required.
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