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Given that a^x = b^y= c^z=d^u and a,b,c,d are in GP show that x,y,z,u are in HP

Given that a^x = b^y= c^z=d^u and a,b,c,d are in GP show that x,y,z,u are in HP

Grade:11

1 Answers

Arun
25758 Points
5 years ago
ax = by = cz = du = k (let)
hence
a = k1/x, b = k1/y , c = k1/z , d = k1/u
now a,b,c,d are in g.p. (given)
hence b/a = d/c
 
k1/y / k1/x  = k1/u / k1/z 
k(1/y –1/x) = k(1/u –1/z)
hence 1/y – 1/x = 1/u –1/z
hence 1/x, 1/y, 1/z, 1/u are in A.P.
hence x, y, z, u will be in H.P.
 
hope it helps

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