Question

For what value of p, are (2p – 1), 7 and 11/2p three consecutive terms of an AP?

Answer 1

Answer:

let's say

a(n)=2p-1

a(n+1)=7

a(n+2)=11/2p

we know,

a(n+1)-a(n)=d.............(¡)

a(n+2)-a(n+1)=d...........(¡¡)

from ¡ and ¡¡

7-2p+1=11/2p-7

7+7=11/2p+2p

14=11/2p+4/2p

14=15/2p

p=14×2/15

p=28/15

Answer..!