A) \[2l=m+n\]
B) \[2m=n+l\]
C) \[m=n+l\]
D) \[l=m+n\]
Correct Answer: B
Solution :
Roots are equal so \[{{b}^{2}}-4ac=0\] Þ \[{{(n-l)}^{2}}-4(m-n)(l-m)=0\] Þ \[{{n}^{2}}+{{l}^{2}}-2nl-4\,(ml-nl-{{m}^{2}}+mn)=0\] Þ \[{{n}^{2}}+{{l}^{2}}-2nl-4ml+4nl+4{{m}^{2}}-4mn=0\] Þ \[{{l}^{2}}+{{n}^{2}}+{{(2m)}^{2}}+2nl-4mn-4ml=0\] Þ \[{{(l+n-2m)}^{2}}=0\] Þ \[l+n=2m\] Þ l, m, n are in A.P.You need to login to perform this action.
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