Question

If k+1,3k and 4k+2 aer any three consecutive terms of ap ,find the value of k

Answer 1

Answer:

Step-by-step explanation:

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1st term t1 = (k + 1).

2nd term t2 = 3k.

3rd term t3 = (4k + 2).

Given that t1,t2,t3 are in AP.

We know that when they are in AP, their common difference will be:

t2 - t1 = t3 - t2

3k - (k + 1) = (4k + 2) - 3k

3k - k - 1 = 4k + 2 - 3k

2k - 1 = k + 2

2k = k + 2 + 1

2k = k + 3

2k - k = 3

k = 3.

Therefore the value of k = 3.

Hope this helps

Answer 2

Hi

Here is your answer

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a1 = (k+1)

a2 = 3k

a3 = (4k + 2)

These are three consecutive terms.

Since,

a2 - a1 = a3 - a2

Therefore,

3k - (k+1) = (4k + 2) - 3k

3k - k - 1 = 4k + 2 - 3k

2k - 1 = k + 2

Bring variables and constants to opposite sides,

2k - k = 2+1

k = 3

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Hope it helps you..