Question

Find the four numbers in AP whose sum is 20 and their product is 384.

Answer 1

Let First term of A.P. = a
And
Common difference = 2d
So
Four numbers in A.P. = ( a - 3d ) , ( a - d ) , ( a + d ) and ( a + 3d )

So, given

⇒( a - 3d ) + ( a - d ) + ( a + d ) + ( a + 3d ) = 20

⇒4a = 20

⇒a = 5

And

⇒( a - 3d ) ( a - d ) ( a + d ) ( a + 3d ) = 384

⇒( a - 3d ) ( a + 3d ) ( a - d ) ( a + d ) = 384

⇒( a2 - 9d2 ) ( a2 - d2 ) = 384

⇒a4 - a2d2 - 9a2d2 + 9d4 = 384

⇒a4 - 10a2d2 + 9d4 = 384 , Now we substitute a = 5 ,and get

⇒54 - 10 × 52 ×d2 + 9d4 = 384

⇒625- 250d2 + 9d4 = 384

⇒9d4 - 250d2 + 241 = 0

from , Splitting the middle term method we get

⇒9d4 - 9d2 - 241d2 + 241 = 0

⇒9d2 ( d2 - 1 ) - 241 ( d2 - 1 ) = 0

⇒( 9d2 - 241 ) ( d2 - 1 ) = 0

So,
d = ±1
And




We take d = 1 or - 1 to get natural numbers of our A.P.

Our four terms of A.P. are

⇒5 - 3× 1 = 2

⇒5 - 1 = 4

⇒5 + 1 = 6

and
⇒5 + 3 × 1 = 8

SO,

Four terms of A.P. that gives, their sum is 20 and product is 384 = 2 , 4 , 6 , 8