# In an AP, the 6th term is half the 4th term and the 3rd term is 15. how many terms are needed to give a sum that is equal to 66?

2
by taemin

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by taemin

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You will get an equation n² - 15n + 44 = 0. n = 11 or n = 4.

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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/q

(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