Question

If K +1 , 3K and 4K +2 be any three consecutive terms of an AP , find the value of K.

Answer 1

Given (k + 1), 3k and (4k + 2) are in AP
Common difference = t2 – t1 = t3 – t2
That is 3k – (k + 1) = (4k + 2) – 3k
⇒ 3k – k – 1 = 4k + 2 – 3k
⇒ 2k – 1 = k + 2
⇒ 2k – k = 1 + 2
∴ k = 3


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Answer 2

As we know that-k
a+c÷2=b
Therefore,
(k+1)+(4k+2)÷2=3k
k+1+4k+2=2×3k
5k+3=6k
6k-5k=3
k=3


Therefore, k=3

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