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You will get an equation n² - 15n + 44 = 0. n = 11 or n = 4.
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Easiest way is to write the terms in terms of a and d
6th term= a+ (6-1)d =a +5d
4th term = a +(4-1)d =a +3d
3rd term = a +(3-1)d =a +2d =15
(a +5d) = (a+3d)/2
⇒2(a +5d) = (a+3d)
⇒2a -a =3d -10d
⇒a =(-7d)
putting this in 3rd term.
 a +2d =15
⇒-7d +2d =15
⇒-5d =15
⇒d =(-3)
so, a = 21
now, sum of n terms = 66
(n/2)(2a +(n-1)d)= 66
⇒n(42 +(n-1)(-3)) =132
⇒45n -3n² =132
⇒n² -15n +44=0
⇒n² -11n -4n +44=0 
⇒n(n-11) -4(n-11)=0
⇒(n-11)(n-4) =0
either n=11 or n =4
after 4th term the sum of all terms will be zero.

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