Insert three harmonic means between 5 and 6.

1.	Insert three harmonic means between 5 and 6.
Answer» 3 harmonic means between 5 and 6 ⇒ 3 arithmetic means between \(\frac{1}{5}\) and \(\frac{1}{6}\). Let A₁, A₂, A₃ be the arithmetic means between \(\frac{1}{5}\) and \(\frac{1}{6}\). Then, \(\frac{1}{5}\), A₁ , A₂ A₃, \(\frac{1}{6}\) from an A.P, Where t₁ = a = \(\frac{1}{5}\), t₅ = a + 4d = \(\frac{1}{6}\) ∴ \(\frac{1}{5}\)+ 4d = \(\frac{1}{6}\) ⇒ 4d = \(\frac{1}{6}\) - \(\frac{1}{5}\) = \(-\frac{1}{30}\) ⇒ d = \(-\frac{1}{120}\) ∴ A₁ = a + d = \(\frac{1}{5}\) + \(\bigg(-\frac{1}{120}\bigg)\) = \(\frac{23}{120}\) A₂ = a + 2d = \(\frac{1}{5}\) + 2 x\(\bigg(-\frac{1}{120}\bigg)\) = \(\frac{1}{5}\) - \(\frac{1}{60}\) = \(\frac{11}{60}\) A₃ = a + 3d = \(\frac{1}{5}\) + 3 x\(\bigg(-\frac{1}{120}\bigg)\) = \(\frac{1}{5}\) - \(\frac{1}{40}\) = \(\frac{7}{40}\) ∴ Required harmonic means are \(\frac{120}{23}\),\(\frac{60}{11}\), \(\frac{40}{7}\).

Answer»

3 harmonic means between 5 and 6

⇒ 3 arithmetic means between \(\frac{1}{5}\) and \(\frac{1}{6}\).

Let A₁, A₂, A₃ be the arithmetic means between \(\frac{1}{5}\) and \(\frac{1}{6}\).

Then, \(\frac{1}{5}\), A₁ , A₂ A₃, \(\frac{1}{6}\) from an A.P,

Where t₁ = a = \(\frac{1}{5}\), t₅ = a + 4d = \(\frac{1}{6}\)

∴ \(\frac{1}{5}\)+ 4d = \(\frac{1}{6}\) ⇒ 4d = \(\frac{1}{6}\) - \(\frac{1}{5}\) = \(-\frac{1}{30}\) ⇒ d = \(-\frac{1}{120}\)

∴ A₁ = a + d = \(\frac{1}{5}\) + \(\bigg(-\frac{1}{120}\bigg)\) = \(\frac{23}{120}\)

A₂ = a + 2d = \(\frac{1}{5}\) + 2 x\(\bigg(-\frac{1}{120}\bigg)\) = \(\frac{1}{5}\) - \(\frac{1}{60}\) = \(\frac{11}{60}\)

A₃ = a + 3d = \(\frac{1}{5}\) + 3 x\(\bigg(-\frac{1}{120}\bigg)\) = \(\frac{1}{5}\) - \(\frac{1}{40}\) = \(\frac{7}{40}\)

∴ Required harmonic means are \(\frac{120}{23}\),\(\frac{60}{11}\), \(\frac{40}{7}\).

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