Question

If 2k+1, 13, 5k-3 are three consecutive terms of AP, then k=? ​

Answer 1

Step-by-step explanation:

here is your answer mate

Answer 2

Answer:

k=4

Step-by-step explanation:

let t1=2k+1 and t2=13 and t3=5k-3

common difference is t2-t1 =t3-t2

13-(2k+1)=5k-3-(13)

13-2k-1=5k-3-13

13+13-1+3=5k+2k

28=7k

k=28/7

k=4

therefore k=4