For what value of k ,x,2x+k and 3x+6 are consecutive terms of A.P
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1
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himanilak:
on the RHS you have taken (3k+6)-(2n+k)
which should be equal to 3k+6-2n-k and not 3k+6-2n+k
So therefore the correct answer will be 3
Answered by
4
(3x + 6) - (2x + k) = (2x + k) - x
3x + 6 - 2x - k = 2x + k - x
x + 6 - k = x + k
6 = 2k
k = 3
AP - x, 2x+3, 3x+6
d= x+3
3x + 6 - 2x - k = 2x + k - x
x + 6 - k = x + k
6 = 2k
k = 3
AP - x, 2x+3, 3x+6
d= x+3
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