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

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Arun · Answer

a x = b y = c z = d u = k (let) hence a = k 1/x , b = k 1/y , c = k 1/z , d = k 1/u now a,b,c,d are in g.p. (given) hence b/a = d/c k 1/y / k 1/x = k 1/u / k 1/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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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

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