How many term of the AP -9,-6,-3....must be taken so that their sum is 66
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a = -9
d = 3
Sn =1/2 n (2a +(n-1)d)
66 = 1/2 n(2(-9) + (n-1)3)
66 = 1/2n (-18+3n-3)
66 = 1/2n(-21+3n)
0 =3/2n²-21/2n -66
0 = 3n²-21n-132
0 = (n-11)(3n+12)
n = 12 n=-4
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