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`G_(k) = (a_(1) a_(2) …. A_(k))^(1//k)`
`implies G_(k) = a_(1) (r^(1 + 2 + …. + (k - 1)))^(1//k)`
`implies G_(k) = a_(1) r^((k - 1)/(2))`
`A_(k) = (a_(1) + a_(2) + … + a_(k))/(k)`
`implies A_(k) = (a_(1) (1 + r + …+ r^(k - 1)))/(K)`
`implies A_(k) = (a_(1)(k^(k) - 1))/((r - 1).K)`
`H_(k) = (k)/((1)/(a_(1)) + (1)/(a_(2)) + .... + (1)/(a_(k)))`
`implies H_(k) = (a_(1)k)/(a + (1)/(r ) + .. + (1)/(r^(k-1))) = (a_(1) k)/((1-(1)/(r^(k))/(1-(1)/(r)))`
`implies H_(k) = (a_(1) k (r - 1) r^(k - 1))/(r^(k) - 1)`
From (1),(2) and (3) we get
`G_(k) = (A_(k) H_(k))^(1//2)`
`implies underset(k = 1)overset(n)(prod) G_(k) = underset(k = 1)overset(n)(prod) (A_(k) H_(k))^((1)/(2))`
`implies (underset(k = 1)overset(n)(prod) G_(k))^(1//n) = (A_(1) A_(2) .... A_(n) xx H_(1) H_(2)..... H_(n))^(1//2n)`
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CENGAGE ENGLISH-INEQUALITIES INVOLVING MEANS -Illustration
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