Question

Determine k so that k+2, 4k-6 and 3k-2 are three consecutive terms of A.P.

Answer 1

To be terms of an AP the difference between two consecutive terms must be the same
So if k+2, 4k-6 & 3k-2 are terms of an AP, then
4k-6–(k+2) = 3k-2–(4k-6)
4k-6-k-2=3k-2-4k+6
3k-8= –k+4
4k = 12
k =12/4=3

Thus the value of k is 3

HOPE IT HELPS..... :-)

Answer 2

Hi friend !

If k+2, 4k-6 and 3k-2 are in A.P , then they would have the same common difference,

4k - 6 - [ k + 2 ] = 3k - 2 - [ 4k - 6]

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

3k - 8 = -k + 4

3k + k = 4 + 8

4k = 12

k = 12/4

k = 3

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You can verify if the no:s are in A.P , if k = 3

k + 2 = 5
4k - 6 = 6
3k - 2 = 7

5,6,7 .... are in A.P

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Therefore ,
k = 3