Determine k so that (3k − 2), (4k − 6) and (k + 2) are three consecutive terms of an AP. [CBSE 2009C]

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Solution

It is given that (3k − 2), (4k − 6) and (k + 2) are three consecutive terms of an AP.

∴ (4k − 6) − (3k − 2) = (k + 2) − (4k − 6)

⇒ 4k − 6 − 3k + 2 = k + 2 − 4k + 6

⇒ k − 4 = −3k + 8

⇒ k + 3k = 8 + 4

⇒ 4k = 12

⇒ k = 3

Hence, the value of k is 3.

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Similar questions

Determine k so that (3k -2), (4k-6) and (k +2) are three consecutive terms of an AP.

k

3 k - 2

4 k - 6

k + 2

A P .

k^{2} + 4 k + 8 = 0, 2 k^{2} + 3 k + 6 = 0, 3 k^{2} + 4 k + 4 = 0

(K + 2), (4 K - 6)

(3 K - 6)

k

k^{2} + 4 k + 8

2 k^{2} + 3 k + 6

3 k^{2} + 4 k + 4

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