Question

k-2, 12-k,2k+1 is an A. P.... find the value of k

please answer me I will mark u as the brainest​

Answer 1

Answer:

k = 5

Step-by-step explanation:

( k-2), (12-k) and (2k+1) are in AP.

So,

Common difference (d) = (12-k) - (k-2)

d = 12 - k - k + 2

d = 14 - 2k ...(I)

Also,

d = (2k+1) - (12-k)

14 - 2k = 2k + 1 - 12 + k

-2k -2k - k = -14 -12 + 1

-5k = -25

5k = 25

On dividing by 5,

k = 5

Thanks, Please mark as brainliest if it helped :D

Answer 2

Answer :-

• Value of k is 5.

Given :-

• Three terms of an AP (k - 2), (12 - k) and (2k + 1)

To Find :-

• Value of k.

Solution :-

Here

• A = k - 2

• B = 12 - k

• C = 2k + 1

→ B - A = C - B

According to question :-

⇒ (12 - k) - (k - 2) = (2k + 1) - (12 - k)

⇒ 12 - k - k + 2 = 2k + 1 - 12 + k

⇒ 12 + 2 - k - k = 2k + k + 1 - 12

⇒ 14 - 2k = 3k - 11

⇒ 14 + 11 = 3k + 2k

⇒ 25 = 5k

⇒ k = 25/5

⇒ k = 5

Put the value of k in the equation.

• A = k - 2 = 5 - 2 = 3
• B = 12 - k = 12 - 5 = 7
• C = 2k + 1 = 2(5) + 1 = 11

Verification :-

→ 11 - 7 = 7 - 3

→ 4 = 4

Hence, verified !