Question

if 2k + 4and3k-7 and k+12 in A.P then k=​

Answer 1

Answer:

So 3k+7, 2k+5, 2k+7 are in AP.

Let’s take them as a1, a2 and a3 respectively.

a2-a1 = a3-a2

(2k+5) - (3k+7) = (2k+7) - (2k+5)

= 2k+5–3k-7 = 2k+7–2k-5

= -k-2 = 2

= -k = 2+2

= -k = 4

∴ k = -4

Answer 2

EXPLANATION.

⇒ 2k + 4 and 3k - 7 and k + 12 are in A.P.

As we know that,

Method = 1.

Common difference of an A.P. = b - a = c - b.

⇒ 3k - 7 - [2k + 4] = k + 12 - [3k - 7].

⇒ 3k - 7 - 2k - 4 = k + 12 - 3k + 7.

⇒ k - 11 = - 2k + 19.

⇒ k + 2k = 19 + 11.

⇒ 3k = 30.

⇒ k = 10.

Method = 2.

Conditions of an A.P.

⇒ 2b = a + c.

⇒ 2(3k - 7) = 2k + 4 + k + 12.

⇒ 6k - 14 = 3k + 16.

⇒ 6k - 3k = 16 + 14.

⇒ 3k = 30.

⇒ k = 10.

                                                                                                                   

MORE INFORMATION.

Supposition of an A.P.

(1) = Three terms as : a - d, a, a + d.

(2) = Four terms as : a - 3d, a - d, a + d, a + 3d.

(3) = Five terms as : a - 2d, a - d, a, a + d, a + 2d.