Question

please give me full solution

Answer 1

Hey


Solution :-

Let the numbers be
( a + 3d ) , ( a + d ) , ( a - d ) , (a - 3d )

Now ,

Sum of these numbers = 28

So ,

a + 3d + a + d + a - d + a - 3d = 28

=> 4a = 28

=> a = 28 / 4

=> a = 7 .

Now also given ,

Sum of thier squares = 216

So ,

( a + 3d ) ² + ( a + d ) ² + ( a - d ) ² + ( a - 3d ) ² = 216


=> ( 7 + 3d ) ² + ( 7 + d ) ² + ( 7 - d ) ² + ( 7 - 3d ) ² = 216


=> 49 + 9d² + 42d + 49 + d² + 14d + 49 + d² - 14d + 49 + 9d² - 42d = 216


=> 20d² + 4 * 49 = 216

=> 20d² + 196 = 216

=> 20d² = 216 - 196

=> 20d² = 20

=> d² = 1

=> d = 1 .

So a = 7 and d = 1

First number = ( 7 + 3 ) = 10

Second number = ( 7 + 1 ) = 8

Third number = ( 7 - 1 ) = 6

Fourth number = ( 7 - 3 ) = 4



thanks :)

Answer 2

Given that the four numbers are in AP.

Let the four numbers be (a-3d),(a+d),(a+3d),(a-d).

Given that Sum of four numbers be 28.

a - 3d + a + d + a+ 3d + a - d = 28

4a = 28

a = 7.


Given that sum of whose square is 216.

(a - 3d)^2 + (a + d)^2 + (a + 3d)^2 + (a - d)^2 = 216

a^2 + 9d^2 - 6ad + a^2 + d^2 + 2ad + a^2 + 9d^2 + 6ad + a^2 + d^2 - 2ad = 216

4a^2 + 20d^2 = 216

a^2 + 5d = 54

7^2 + 5d = 54

49 + 5d = 54

5d = 54 - 49

5d = 5

d = +1 (or) -1.


When a = 7 and d = +1.

a + 3d = 7 + 3(1)

            = 10.


a + d = 7 + 1

          = 8


a - d = 7 - 1

         = 6


a - 3d = 7 - 3(1)

           = 4


When a = 7 and d = -1.


a + 3d = 7 + 3(-1)

            = 4


a + d = 7 + (-1)
          
          = 6


a - d = 7 - (-1)

         = 8


a - 3d = 7 - 3(-1)

           = 7 + 3

           = 10.




Therefore the four numbers are 4,6,8,10.



Hope this helps!