Math, asked by Faeka265, 1 year ago

2k+1,16 and 9k-2 is A.P k=?

Answers

Answered by Anonymous
3
if these terms are in AP then 16-(2k+1)=9k-2-(16)
why solving we get k is equals. to 3.
Answered by Anonymous
4
2k + 1, 16, 9k - 2 are in A. P.

Then, their common difference will be equal.

16 - ( 2k+1 ) = 9k - 2 - 16

16 - 2k - 1 = 9k - 18

15 - 2k = 9 k - 18

15 + 18 = 9k + 2k

11k = 33

k = 33/ 11

k = 3
Similar questions