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If there consecutive terms of an AP are k+2 ,4k -6 and 3k-2 . What is the value of k​

User LynxZh
by
4.7k points

1 Answer

5 votes

Answer: k = 3

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Step-by-step explanation:

  • A = first term = k+2
  • B = second term = 4k-6
  • C = third term = 3k-2

To go from the first term to the second term, we add on some common difference d.

So,

B = A+d

B = (k+2)+d

4k-6 = k+2+d

4k-6-k-2 = d

d = 3k-8

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Similarly, to go from the second term to the third term, we also add on d

C = B+d

C = (4k-6)+d

C = (4k-6)+(3k-8)

C = 7k-14

3k-2 = 7k-14

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Let's solve for k

3k-2 = 7k-14

-2+14 = 7k-3k

12 = 4k

4k = 12

k = 12/4

k = 3 is the final answer

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If k = 3, then we have these three terms:

  • A = k+2 = 3+2 = 5
  • B = 4k-6 = 4(3)-6 = 6
  • C = 3k-2 = 3(3)-2 = 7

The arithmetic progression (AP) is 5, 6, 7. The common difference is d = 1.

Note how d = 3k-8 = 3(3)-8 = 1

User Goneskiing
by
6.3k points
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