Question

The sum of four numbers in AP is 24 and the product of the squares of the exterme is 729 . Find the numbers

Answer 1

Given:

• Sum of four numbers in AP is 24
• Product of the squares of the extreme is 729

To Find:

All the numbers of given A.P.

Solution:

Let the numbers be a-3d, a-d, a+d, a+3d which are on A.P. with common difference 'd'.

Now,

According to the question

a-3d+a-d+a+d+a+3d= 24

⇒ 4a= 24

⇒ a= 6

Here extreme means first and last term of given A.P. i.e. a-3d and a+3d.

We know that,

• a²b²= (ab)²
• (a+b)(a-b)= a² -b²

So, According to the question

(a-3d)²(a+3d)²= 729

((a-3d)(a+3d))²= 729

(a² -(3d)²)²= 729

a² -(3d)²= √(729)

a² -(3d)²= 27

Now, on putting value of a, we get

(6)²-9d²= 27

36-9d²= 27

-9d²= 27-36

-9d²= -9

d²= 1

d= ±1

So, required terms are

Case 1: When d=1

6-3(1), 6-1, 6+1, 6+3(1)

3, 5, 7, 9

Case 2: When d= -1

6-3(-1), 6-(-1), 6-1, 6+3(-1)

9, 7, 5, 3

Hence, the required terms are 3,5,7,9 or 9,7,5,3.