Insert 6

Insert four harmonic means between 3/7 and 3/19

Correct result:

a₂ =  0.355
a₃ =  0.303
a₄ =  0.2643
a₅ =  0.2344

Solution:

$$\begin{gathered} a_1=3\mathrm{/}7={\displaystyle \frac{3}{7}}\doteq 0.4286 \\ {a}_6=4\mathrm{/}19={\displaystyle \frac{4}{19}}\doteq 0.2105 \\ {b}_1=1\mathrm{/}{a}_1=1\mathrm{/}0.4286={\displaystyle \frac{7}{3}}=2\frac{1}{3}\doteq 2.3333 \\ {b}_6=1\mathrm{/}{a}_6=1\mathrm{/}0.2105={\displaystyle \frac{19}{4}}=4\frac{3}{4}=4.75 \\ d=(b_6-b_1)\mathrm{/}(6-1)=(4.75-2.3333)\mathrm{/}(6-1)={\displaystyle \frac{29}{60}}\doteq 0.4833 \\ {b}_2=b_1+d=2.3333+0.4833={\displaystyle \frac{169}{60}}=2\frac{49}{60}\doteq 2.8167 \\ {a}_2=1\mathrm{/}{b}_2=1\mathrm{/}2.8167={\displaystyle \frac{60}{169}}=0.355 \end{gathered}$$

$$\begin{gathered} b_3=b_2+d=2.8167+0.4833={\displaystyle \frac{33}{10}}=3\frac{3}{10}=3.3 \\ {a}_3=1\mathrm{/}{b}_3=1\mathrm{/}3.3={\displaystyle \frac{10}{33}}=0.303 \end{gathered}$$

$$\begin{gathered} b_4=b_3+d=3.3+0.4833={\displaystyle \frac{227}{60}}=3\frac{47}{60}\doteq 3.7833 \\ {a}_4=1\mathrm{/}{b}_4=1\mathrm{/}3.7833={\displaystyle \frac{60}{227}}=0.2643 \end{gathered}$$

$$\begin{gathered} b_5=b_4+d=3.7833+0.4833={\displaystyle \frac{64}{15}}=4\frac{4}{15}\doteq 4.2667 \\ {a}_5=1\mathrm{/}{b}_5=1\mathrm{/}4.2667={\displaystyle \frac{15}{64}}=0.2344 \end{gathered}$$

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