How many terms of the ap 18,16,4 .... Be taken so that their sum is zero?
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a=18
d=16-18= -2
Sn=0
Sn=n/2{2a+(n-1)d}
0=n/2{2×18+(n-1)(-2)}
0=n/2{36+(n-1)(-2)}
0=n/2(36-2n+2)
0=n/2(36+2-2n)
0=n/2(38-2n)
0×2=n(38-2n)
0=n(38-2n)
0=38n-2n²
2n(n-19) = 0
n-19=0×2n
n-19=0
n=19
d=16-18= -2
Sn=0
Sn=n/2{2a+(n-1)d}
0=n/2{2×18+(n-1)(-2)}
0=n/2{36+(n-1)(-2)}
0=n/2(36-2n+2)
0=n/2(36+2-2n)
0=n/2(38-2n)
0×2=n(38-2n)
0=n(38-2n)
0=38n-2n²
2n(n-19) = 0
n-19=0×2n
n-19=0
n=19
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