Question

Determine the value of k so that k + 2 and 4 k - 6 and 3 k -2 are three consecutive terms of an arithmetic progression.

Answer 1

ANSWER:

k = 3

STEP BY STEP EXPLANATION:

Let the first term a = k+2

Then second term b = 4k-6

And third term c = 3k-2

Let the common difference be d

We know that difference between the second term and the first term is d

So , d = 4k-6-(k+2) = 3k-8------------> eq(1)

Also the difference between third term and second term is d because all are in a.p where difference is common

So , d = 3k-2-(4k-6) = -k+4----------> eq(2)

Eq(1) = Eq(2) = d

3k-8 = -k+4

3k+k = 8+4

4k = 12

k = 3

Hope it helps

Pls mark as brainliest

Thanks