Question 251659: Q1 If a,b,c,d are in Harmonic progression show that(a-c)(b-d)=4(a-b)(c-d)
Q2 If a+b,b+c,c+a are in H.P. show that a^2,b^2,c^2 are in A.P.
Q3 If x,y,z are in A.P.x,xy,z are in G.P.show thatx,x^2y,z are in H.P.
Answer by palanisamy(496)   (Show Source):
You can put this solution on YOUR website! Q1.
a,b,c,d are in H.P
Then 1/a, 1/b, 1/c, 1/d will be in A.P
Therefore 2(1/b-1/a) = (1/c-1/a) and 2(1/d-1/c) = (1/d-1/b)
2(a-b)/ba = (a-c)/ca and 2(c-d)/dc = (b-d)/db
Multiplying these two equations, we get
4(a-b)(c-d)/abcd = (a-c)(b-d)/abcd
4(a-b)(c-d) = (a-c)(b-d)
Q2.
Given a+b,b+c,c+a are in H.P.
Therefore, 1/(a+b) + 1/(c+a) = 2/(b+c)
(c+a+a+b)/(a+b)(c+a) = 2/(b+c)
(c+2a+b)(b+c) = 2(a+b)(c+a)
bc+c^2+2ab+2ac+b^2+bc = 2ac+2a^2+2bc+2ab
c^2+b^2 = 2a^2
(b^2+c^2)/2 =a^2
Therefore b^2,a^2,c^2 are in A.P
Q3 Given, x,y,z are in A.P.
Therefore, x+z = 2y ......(1)
Also, x,xy,z are in G.P
Therefore, (xy)^2 = xz
x^2y^2 = xz
xy^2 = z ....(2)
Next, we will prove x,x^2y,z are in H.P
Now, 1/x+1/z = (z+x)/xz
= 2y/x.xy^2
= 2/x^2y
Therefore, 1/x+1/z =2/x^2y
Therefore,x,x^2y,z are in H.P
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