Harmonic series

Insert four members between 5/3 and 5/11 to form harmonic series (means).

Correct answer:

a₂ =  1.087
a₃ =  0.8065
a₄ =  0.641
a₅ =  0.5319

Step-by-step explanation:

$$\begin{gathered} a_1=5\mathrm{/}3={\displaystyle \frac{5}{3}}=1\frac{2}{3}\doteq 1.6667 \\ {a}_6=5\mathrm{/}11={\displaystyle \frac{5}{11}}\doteq 0.4545 \\ {b}_1=1\mathrm{/}{a}_1=1\mathrm{/}1.6667={\displaystyle \frac{3}{5}}=0.6 \\ {b}_6=1\mathrm{/}{a}_6=1\mathrm{/}0.4545={\displaystyle \frac{11}{5}}=2\frac{1}{5}=2.2 \\ d=(b_6-b_1)\mathrm{/}(6-1)=(2.2-0.6)\mathrm{/}(6-1)={\displaystyle \frac{8}{25}}=0.32 \\ {b}_2=b_1+d=0.6+0.32={\displaystyle \frac{23}{25}}=0.92 \\ {a}_2=1\mathrm{/}{b}_2=1\mathrm{/}0.92={\displaystyle \frac{25}{23}}=1\frac{2}{23}=1.087 \end{gathered}$$

$$\begin{gathered} b_3=b_2+d=0.92+0.32={\displaystyle \frac{31}{25}}=1\frac{6}{25}=1.24 \\ {a}_3=1\mathrm{/}{b}_3=1\mathrm{/}1.24={\displaystyle \frac{25}{31}}=0.8065 \end{gathered}$$

$$\begin{gathered} b_4=b_3+d=1.24+0.32={\displaystyle \frac{39}{25}}=1\frac{14}{25}=1.56 \\ {a}_4=1\mathrm{/}{b}_4=1\mathrm{/}1.56={\displaystyle \frac{25}{39}}=0.641 \end{gathered}$$

$$\begin{gathered} b_5=b_4+d=1.56+0.32={\displaystyle \frac{47}{25}}=1\frac{22}{25}=1.88 \\ {a}_5=1\mathrm{/}{b}_5=1\mathrm{/}1.88={\displaystyle \frac{25}{47}}=0.5319 \end{gathered}$$

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