Question

Find the value of 'K' so that 5k + 3, 6k + 3, 8k - 4 will form three consecutive
terms of an A.P.​

Answer 1

Answer:

a,b,c are three consecutive terms of an A.P then 2b=a+c

So, 2(2k2+3k+6)=k2+4k+8+3k2+4k+4

4k2+6k+12=4k2+8k+12

6k=8k

∴k=0

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Answer 2

Answer :-

• Value of k is 7.

Given :-

• Three terms of an AP (5k + 3), (6k + 3) and (8k - 4).

To Find :-

• Value of k.

Solution :-

Here

Three terms of an AP are

• (5k + 3)
• (6k + 3)
• (8k - 4)

As we know that

• C - B = B - A

Let

• A = (5k + 3)
• B = (6k + 3)
• C = (8k - 4)

Now

⇒ (8k - 4) - (6k + 3) = (6k + 3) - (5k + 3)

⇒ 8k - 4 - 6k - 3 = 6k + 3 - 5k - 3

⇒ 8k - 6k - 4 - 3 = 6k - 5k + 3 - 3

⇒ 2k - 7 = k + 0

⇒ 2k - 7 = k

⇒ 2k - k = 7

⇒ k = 7

Verification :-

⇒ (8k - 4) - (6k + 3) = (6k + 3) - (5k + 3)

⇒ 8(7) - 4 - 6(7) - 3 = 6(7) + 3 - 5(7) - 3

⇒ 56 - 4 - 42 - 3 = 42 + 3 - 35 - 3

⇒ 56 - 4 - 45 = 42 + 3 - 38

⇒ 56 - 49 = 42 - 35

⇒ 7 = 7

Hence, verified !