Question

In an AP, 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?
PLEASE GIVE THE ANSWER STEP BY STEP.

Answer 1

here is the answer for the question
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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 [2 a + ( 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 .