Question

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

Answer 1

Let the common difference of the AP be d

Therefore,  According to the problem,

(k+3)+d = 4k-6                         ...(1)

and                                          

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

Now on observing the equation, we find that these are pair of linear equations in two variables.

Now we solve them by eliminating d from them

Subtracting (1) from (2)

4k - 6 + d - k - 3 - d = 4k - 6 - 3k + 3

3k -9 =  -3

So k = 2

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Answer 2

I am getting the ans 3 but it is not satisfying the three consecutive terms in an AP

sry, either my ans is wrong or your question is wrong