In an AP,l^th,m^th,n^th terms are 2,4,6 respectively.The equation lx²-2mx+n=0 has

a.one positive and one negative root

b.both roots negative

c.imaginary roots

d.both roots positive

Hi Menka,

As 2,4,6 are the lth,mth, and nth term. We can write

2 = a+(l-1)d

4 = a+(m-1)d

6 = a+(n-1)d

So 4-2 = (m-l)d, 6-4 = (n-m)d

or m-l = n-m, or 2m=l+n.

Now discriminant of the QE, D = 4m^2 - 4ln = (l+n)^2 - 4ln = (l-n)^2, which is always positive, and hence roots are real.

Now roots = [2m ± (l-n)]/(2l) = [l+n ±(l-n)]/2l = 1, n/l

As l,m,n are positive (as they refer the term number of an AP)

So roots are always positive.

Option (D).

Hope this helps.

Best Regards,

Ashwin (IIT Madras).