Question

50 Points!! Only Answer Of You Know. Otherwise I Will Report And Your Answer Will Be Deleted And You Wont Get Any Point.

Question:50 Points!! Only Answer Of You Know. Otherwise I Will Report And Your Answer Will Be Deleted And You Wont Get Any Point.Instead A Warn!

Question:Determine k so that k+2, 4k - 6 and 3k - 2 are the three consecutive terms of an AP.
​​

Answer 1

$\huge\mathbb\green{Answer:-}$

For consecutive terms of an AP,

a₃ - a₂ = a₂ - a₁

Let a₁ = k + 2, a₂ = 4k - 6 & a₃ = 3k - 2.

Putting the values :-

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

⇒ 3k - 2 - 4k + 6 = 4k - 6 - k - 2

⇒ -2 + 6 + 6 + 2 = 4k - k - 3k + 4k

⇒ 12 = 4k

⇒ k = 12/4

⇒ k = 3

So, for k = 3 k+2, 4k - 6 and 3k - 2 are the three consecutive terms of an AP.

Answer 2

If three terms are in AP, the difference between the terms should be equal, i.e. if a, b and c are in AP then,

b - a = c - b

Since, the terms are in an AP, therefore

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

⇒ k - 4 = - 3k + 8

⇒ 4k = 12

⇒ k = 3

∴ k = 3