Question

If k,2k-1 and 2k+1 are three consecutive terms of an AP then the value of K is

Answer 1

hence k=3 hope this helped u plzz mark as brainliest

Answer 2

As we know,

General form of A.P=....a-d,a,a+d...

Given: a-d=k.......1

a=2k-1 .......2

a+d=2k+1.........3

Let's add them..

:(a-d)+a+(a+d)=k+(2k-1)+(2k+1)

: positive and negative d at lhs and 1 at rhs got cancelled.

:3a=5k+0

Now, by elimination method..

Multiply three in .....2

3×(a=2k-1)

3a=6k-3

Eliminating,

3a=6k-3

-(3a=5k+0)

:0=k-3

:K=3

The value of k is 3..

Hope ! This is helpful....!;!!!