Question

Determine k, so that k+2, 4k -6 and 3k -2 are there consecutive terms of an
a.P.

Answer 1

(4k-6)-(k+2)=(3k-2)-(4k-6)

4k-6-k-2 = 3k-2-4k+6

4k = 4+8

4k = 12

k = 3

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Answer 2

Answer:

$k=3$

Step-by-step explanation:

We have given in a question that there are 3 consecutive terms of an A.P

We know that in that difference of 1st and 2nd term= difference of 2nd and 3rd term of an A.P

By this formula we can get the value of k

So the equation might be

$4k-6-k-2=3k-2-4k+6$

=$3k-8=4-k$

=$3k+k=4+8$

$4k=12$

So our final answer might be

$k=3$