Question

For what value of k,k 9,2,k-1 and 2k 7 are the consecutive term of an AP

Answer 1

Let,

k + 9 = a

2k - 1 = b

2k + 7 = c

 

To be in AP,

 

a + c = 2b

(k + 9) + (2k + 7) = 2(2k - 1)

k + 9 + 2k + 7 = 4k - 2

3k + 16 = 4k - 2

3k - 4k = - 2 - 16

 - k = - 18

k = 18

 

For k = 18, the terms k+9 , 2k - 1 , 2k + 7 are in AP

 

There’s an alternate method too:

k+9, 2k-1,2k+7. It is an A.P

let First ,second & third terms be t1,t2,t3.

The difference between the two terms is constant.

t2–t1 = t3-t2

(2k-1)-(k+9) = (2k+7)-(2k-1)

2k-1-k-9 = 2k +7 - 2k +1

2k-k -1–9 = 2k -2k +7+1

k-10 = 8

k=8+10

K = 18

The numbers are 18+9, 36–1, 36+7

= 27, 35 & 43 . The common difference is 8.

Ans : 18

Hope This Helps :)

Answer 2

Answer:

hi!!

k+9, 2k-1,2k+7. It is an A.P

let First ,second & third terms be t1,t2,t3.

The difference between the two terms is constant.

t2–t1 = t3-t2

(2k-1)-(k+9) = (2k+7)-(2k-1)

2k-1-k-9 = 2k +7 - 2k +1

2k-k -1–9 = 2k -2k +7+1

k-10 = 8

k=8+10

K = 18

The numbers are 18+9, 36–1, 36+7

= 27, 35 & 43 . The common difference is 8.

Ans : 18

hope this helps

please mark my answer as the brainliest

thanks