find four natural numbers in ap such that their sum is 20 and product is 384

Hi Sadia Hafeez,

Kindly find the solution to your query:

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
d = ± 241√3

But as our numbers are natural numbers , So 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 

To practice with some more questions following are the links of the similar queries as asked by you:

https://www.meritnation.com/ask-answer/question/three-positive-integers-a1-a2-a3-are-in-ap-such-that-a1/arithmetic-progressions/1502140

https://www.meritnation.com/ask-answer/question/q1-find-the-3-members-of-an-ap-whose-sum-is-20-and-product/math/3543750

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