Question

How many terms of the series -9,-6,-3.. Must taken that sum may be 66?

Answer 1

d = +3

Sn=66

a= -9

Sn=n/2(2a+(n-1)d)

Substituting the values we get,

66=n/2(-18+(n-1)3)

66=n/2(-18+3n-3)

132=n(-21+3n)

132=3n(-7+n)

44=-21n+3n2

3n2-21n-44=0

Then find n by solving it.