Question

18. For what value of k, 2k-1,7 and 3k will form an A.P​

Answer 1

Step-by-step explanation:

Given: k, 2k - 1, 7 and 3k are in AP.

Therefore, their common difference will be equal.

Hence ,

$\therefore \: 2k - 1 - k = 3k - 7 \\ \therefore \: k - 1 = 3k - 7 \\ \therefore \: k - 3k = 1 - 7 \\ \therefore \: - 2k = - 6 \\ \therefore \: k = \frac{ - 6}{ - 2} \\ \therefore \: \huge \fbox{k =3}$

Thus, for k = 3, given terms will be in AP.

Answer 2

Answer:

k 3

Step-by-step explanation:

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