Question

the sum of first 3 terms of an AP is half the sum of next 3 terms.If the first term is 6.Find the Common difference.

Answer 1

common difference is 3

Answer 2

Here is your solution :

Let the common difference is d.

Given,

First term = 6

So,

A.P. = 6 , 6 + d , 6 + 2d , 6 + 3d , 6 + 4d , 6 + 5d.

Now,

=> Sum of First 3 terms = ( 1/2 ) [ Sum of next 3 terms ]

=> ( 6 + 6 + d + 6 + 2d ) = ( 1/2 ) [ 6 + 3d + 6 + 4d + 6 + 5d ]

=> ( 18 + 3d ) = ( 1/2 ) [ 18 + 12d ]

=> 2( 18 + 3d ) = ( 18 + 12d )

=> 36 + 6d = 18 + 12d

=> 36 - 18 = 12d - 6d

=> 18 = 6d

=> d = 18 ÷ 6

•°• d = 3

Hence, the common difference is 3.

Another method ,

From that method, we could solve easy questions but imagine there are big numbers , so it is not gonna to work.

Here we go for another method !

Given,

First term ( a ) = 6

Let , common difference is d.

=> Sum of n terms = ( n/2 ) [ 2a + ( n - 1 ) d ]

=> Sum of 3 terms = ( 3/2 ) [ 2× 6 + ( 3 - 1 )d ]

=> Sum of 3 terms = ( 3/2 ) [ 12 + 2d ] -------- ( 1 )

Now, for next three terms.

First term will be the fourth term of initial A.P.

=> nth term = a + ( n - 1 )d

=> 4th term = a + ( 4 - 1 )d

=> 4th term = a + 3d

Here, this 4th term is the first term of another A.P.

First term = a + 3d

Common difference = d

=> Sum of n term = ( n/2 ) [ 2a + ( n - 1 ) d ]

=> Sum of 3 terms = ( 3/2 ) [ 2( 6 + 3d ) + ( 3 - 1 )d ]

=> Sum of 3 term = ( 3/2 ) [ 12 + 6d + 2d ]

=> Sum of 3 terms = ( 3/2 ) [ 12 + 8d ] -- ( 2 ) ---- ( 2 )

A/Q,

=> ( 1 ) = ( 2 ) / 2

=> ( 3/2 ) [ 12 + 2d ] = ( 3/2 ) [ 12 + 8d ] ( 1/2 )

=> 2( 12 + 2d ) = 12 + 8d

=> 24 + 4d = 12 + 8d

=> 24 - 12 = 8d - 4d

=> 12 = 4d

=> d = 12 / 4

•°• d = 3

Hope it helps !! ^_^