Question

Q. for what value of k will k+9, 2k-1 and 2k+7 are the consecutive terms of an

a.p.

Answer 1

Hi...☺

Here is your answer...✌

We know that

If a , b , c are in AP

∴ 2b = a + c

Now
Given consecutive term of AP are

k + 9 , 2k - 1 , 2k + 7

∴ 2(2k - 1) = k + 9 + 2k + 7

4k - 2 = 3k + 16

4k - 3k = 16 + 2

⇒ k = 18

Answer 2

Hiii friend,

AP = K+9 , 2K-1 , 2K+7

T1 = K+9

T2 = 2K-1

T3 = 2K+7


First term = K+9

Common difference (D1) = T2-T1

=> 2K-1 - (K+9)

=> 2K -1 -K -9 = K -10

OR,

Common difference (D2) = T3-T2

=> 2K+7-(2K-1)

=> 2K +7 -2K +1 = 7+1 = 8

As we know that the common difference of an AP will always equal.

So,

D1 = D2


K-10 = 8

K = 8+10

K = 18.

Hence,

The Value of k is 18


HOPE IT WILL HELP YOU...... :-)