Question

Find the value of k for which the following term are the AP.
a) 2k+1,k^2+k+1,3k^3-3k+3

Answer 1

If the given terms are in A. P. Then their common difference will be same.

So, ( k^2 + k + 1 ) - ( 2k + 1 ) = (3 k^2 - 3k + 3 ) - ( k^2 + k + 1 )

=> ( k ^2 + k +1 - 2k - 1 ) = ( 3 k^2 - 3k +3 - k^2 - k - 1 )

=> k^2 - k = 3k^2 - 4k +2 - k^2

=> 2k^2 = 3k^2 - 4k +k + 2

=> 0 = 3 k^2 - 2k^2 - 3k + 2

=> 0 = k^2 - 3k +2

=> 0 = k ^2 - 2k - k + 2

=> 0 = k ( k - 2 ) - 1 ( k - 2 )

=> 0 = ( k - 1 ) ( k - 2 )

=> ( k - 1 ) = 0, ( k - 2 ) = 0

=> k = 1, 2.

So, there are two possible values of k = 1, 2.

CORRECTION IN THE QUESTION :

There should be 3k^2 - 3k + 3 instead of ( 3k^3 - 3k +3 ).