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It is a hell for me to figure-out a solution to this: Given that (2-3k), (2k+3) and (k+7) where k is a constant, are three consecutive terms of a linear sequence(A.P.),find common difference. Please sir, help me out of this knotty problem!!

Take thw 2nd minus the 1st and the result is 2k+3 - (2-3k) = 5k+1.

Take the 3rd minus the 2nd and the result is k+7 - (2k+3) = -k+4.

To be consecutive terms, these differences must be equal.

For 5k+1 = -k+4, adding k-1 to both sides gives 6k = 3.

Dividing both sides by 6 gives k = 1/2.

Trying it gives the terms 1/2, 4, 7 1/2.

As can be seen, 4 - 1/2 = 3 1/2 and 7 1/2 - 4 = 3 1/2,

so the difference is the same.

It could be said the terms were 1/2, 8/2, and 15/2.

Doing it this way makes it obvious each term is 7/2 larger than the last term.

Algebra

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