Answers

2015-04-13T19:01:25+05:30


given,

k , 2k-1 and 2k+1 are in a.p

then,

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

k-1      = 2k+1-2k+1

k-1      =2

k          =2+1

k          =3

therefore k value is 3

0
2015-04-13T19:08:20+05:30
In an A.P d=t2-t1 K,2k-1,2k+1(given) => 2k-1-k=2k+1-(2k-1) =>k-1=2k+1-2k+1 =>k-1=2 =>k=2+1 =>k=3 The value of k is 3
0