Question

For what value of k will the consecutive terms 3k +2, 4k+ 5 and 6k - 4 form an A.P.?​

Answer 1

Answer:

when k = 12, they form an A.P

Step-by-step explanation:

Let's say a1, a2 and a3 are in A.P

Then,

we know that, their common difference will always be equal.

a1 + d = a2

a2 + d = a3

Then,

d = a2 - a1

d = a3 - a2

So,

a2 - a1 = a3 - a2

Then,

To make (3k + 2), (4k + 5) and (6k - 4) their common difference must be equal.

So,

Let,

a1 = (3k + 2)

a2 = (4k + 5)

a3 = (6k - 4)

So,

a2 - a1 = a3 - a2

(4k + 5) - (3k + 2) = (6k - 4) - (4k + 5)

Opening the brackets,

4k + 5 - 3k - 2 = 6k - 4 - 4k - 5

k + 3 = 2k - 9

2k - k = 9 + 3

k = 12

OR

In an A.P,

We know that,

(a1 + a3)/2 = a2

So,

((3k + 2) + (6k - 4))/2 = 4k + 5

(3k + 2 + 6k - 4)/2 = 4k + 5

9k - 2 = 2(4k + 5)

9k - 2 = 8k + 10

9k - 8k = 10 + 2

k = 12

Hence,

when k = 12, they form an A.P.

Hope it helped and believing you understood it........All the best