Question

Find four numbers in ap whose sum is 28 and the sum of whose square is 216

Answer 1

Let numbers are ( a - 3d ) , ( a - d ) , ( a + d ) , ( a + 3d )



Given, their sum = 28

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

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

⇒ 4a = 28

⇒ a = 28 / 4

⇒ a = 7




Given, Sum of their squares = 216

⇒ ( 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 + 9d^2 - 42d + 49 + d^2 - 14d + 49 + d^2 + 14d + 49 + 9d^2 + 42d ) = 216


⇒ [ 4( 49 ) + 2( 9d^2 ) + 2( d^2 ) ] = 216


⇒ 2( 49 ) + 9d^2 + d^2 = 108


⇒ 10d^2 = 108 - 98


⇒ 10d^2 = 10


⇒ d^2 = 10 / 10

⇒ d^2 = 1

⇒ d = + 1 OR - 1




Hence, AP ,

=> ( a - 3d ) = 7 - 3( 1 ) = 7 - 3 = 4

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

=> ( a + d ) = 7 + 1 = 8

=> ( a + 3d ) = 7 + 3( 1 ) = 7 + 3 = 10

AP = 4 , 6 , 8 , 10




OR



AP is ,

( a - 3d ) = 7 - 3( - 1 ) = 7 + 3 = 10

( a - d ) = 7 - ( - 1 ) = 7 + 1 = 8

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

( a + 3d ) = 7 + 3( - 1 ) = 7 - 3 = 4

AP = 10 , 8 , 6 , 4