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Given any three successive terms in an harmonic progression, {...,x, y, z,...}, the following is true

$y=\dfrac { 2xz }{ x+z }$

If $x=\dfrac{3}{11}$ and $z=\dfrac{8}{11}$ , then

$y=\dfrac { 2\dfrac { 3 }{ 11 } \dfrac { 8 }{ 11 } }{ \dfrac { 3 }{ 11 } +\dfrac { 8 }{ 11 } } =\dfrac { 48 }{ 121 }$

which is a rational number, as would be expected for any rational $x, z$ .

Harmonic Progression

As we know that both the terms leaving lamda are rational so lambda has to be rational

Let $a = 0.272727....$ and $b = 0.727272 ....$

$100a = 27.272727 .... \text{ \& } 100b = 72.727272 ....$

\Rightarrow 99a = 27 \text { & } 99b = 72

$a = \dfrac{3}{11} , b = \dfrac{8}{11}$

Now

$\dfrac{3}{11} < \lambda < \dfrac{8}{11}$

$\lambda = \dfrac{5}{11}$

which is rational, hence $\lambda$ need not be irrational.