Determine k so that (3k-2),(4k-6)and(k+2) are three consecutive term of an
a.P
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Answered by
4
Answer:
Step-by-step explanation:
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
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Answered by
3
Answer:
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
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