Question

Find the value of k for which the following terms are in A.P.
a)2k+1,k²+k+1,3k²-3k+3​

Answer 1

hey mate

here is the answer....

Answer:

as the terms are in a.p

thus,

2t2=t1+t3

2(k²+k+1)=(2k+1)+(3k²-3k+3)

2(k²+k+1)=2k+1+3k²-3k+3

2(k²+k+1)=-k+4+3k²

2(k²+k+1)=3k²-k+4

2k²+2k+2=3k²-k+4

k²-3k+2=0

k²-k-2k+2=0

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

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

thus,

k=1 or 2

Answer 2

Solution:-

If the Given terms are in A.P.

Then, there Common Difference (D) should be same.

The Three Given Terms are:-

(2k+1) , (k²+k+1) & (3k²-3k+3)

Let (k²+k+1) be the middle term.

=) (k²+k+1) - (2k+1) = (3k²-3k+3) - (k²+k+1)

=) (k²+k+1) - (2k+1) - (3k²-3k+3) + (k²+k+1) = 0

=) k² + k + 1 - 2k - 1 - 3k² + 3k - 3 + k² + k + 1 = 0

=) -k² + 3k -2 = 0

Multiplying both the sides by (-1).

=) k² - 3k + 2 = 0

=) k² - ( 2 + 1)k + 2 = 0

=) k² - 2k - k + 2 = 0

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

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

=) k = 2 and k =1.

Hence,

The Value of "k" for which the given terms are in A.P is 2 and 1.