find the value of 'k',so that k+2, 4k-6 and 3k-2 are the three consecutive terms of an A P
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Step-by-step explanation:
In this A.P
k+2 , 4k-6 , 3k-2
a = k + 2
d = A2 - A1 = A3 - A2
(4k-6) - (k+2) = (3k-2) - (4k-6)
4k - 6 - k - 2 = 3k - 2 - 4k + 6
3k - 8 = -k + 4
4k = 12
k = 3
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