Question

If 2k-6, 4k-6 and 3k-2 are three consecutive terms of an ap, find k

Answer 1

Solution:

Let a1 = 2k-6, a2= 4k-6, a3=3k-2 are three consecutive terms of an a.p

a2 - a1 = a3 - a2

=> 4k-6 -(2k-6) = 3k-2-(4k-6)

=> 4k-6-2k+6 = 3k-2-4k+6

=> 2k = -k + 4

=> 2k+k = 4

=> 3k = 4

=> k = 4/3

Therefore,

k = 4/3

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