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In an AP,lth,mth,nth 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).
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