Question

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?

Answer 1

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/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
either n=11 or n =4
after 4th term the sum of all terms will be zero.

Answer 2

Hey there!

Given,
6th Term of an A. P is half the fourth term.
3rd term of A. P is 15 .

Let the first term of this Arithmetic progression be a , common difference be d.

We know that, n th term = a + ( n - 1 ) d.

So,
6th term = a + 5d
4th term = a + 3d ,

According to the question,
2( a + 5d ) = a + 3d
2a + 10d = a + 3d
a = -7d

Also,
Third term = 15
a + 2d = 15
-7d + 2d = 15
-5d = 15
d = -3 .

Now, a = -7( -3) = 21 .

Let x terms of the A. P gives the sum 66 .

Sum of x terms = x/2 [ 2a + ( x - 1)d ]

66 = x/2 ( 2*21 + ( x - 1 ) ( -3 )

132 = x ( 42 -3x + 3 )

132 = 3x ( 14 - x + 1 )

44 = x ( - x + 15 )

44 = -x² + 15x

x² - 15x + 44 = 0

x² - 11x -4x + 44 = 0

x( x - 11 ) - 4 ( x -11 ) = 0

( x - 4 ) ( x - 11 ) = 0

Either x = 4 or x = 11 .

Therefore, Either 4 or 11 terms are required to get a sum 66 .