Question

2k+1, 3k and 3k-4 are three
consecutive term of AP Then find k​

Answer 1

$a_1=2k+1\\\\a_2=3k\\\\a_3= 3k-4$

$d=a_2-a_1=a_3-a_2$

$3k-2k+1=3k-4-3k\\-2k+1=-4-3k\\1+4=-3k+2k\\5=-1k\\k=(-5)$

hope it helps u :-)

Answer 2

Step-by-step explanation:

=2k+1

a

2

=3k

a

3

=3k−4

d=a_2-a_1=a_3-a_2d=a

2

−a

1

=a

3

−a

2

\begin{gathered}3k-2k+1=3k-4-3k\\-2k+1=-4-3k\\1+4=-3k+2k\\5=-1k\\k=(-5)\end{gathered}

3k−2k+1=3k−4−3k

−2k+1=−4−3k

1+4=−3k+2k

5=−1k

k=(−5)