Question

Insert six harmonic means between 3 and $\dfrac{6}{23}$. Find 22 times of the third term of the six terms?

Answer 1

Hint: Insert 6 variables between 3 and $\dfrac{6}{23}$. As these are in harmonic progression (H.P), their reciprocal would be in Arithmetic Progression (A.P). So use the formula for ${{n}^{th}}$ term of A.P is ${{a}_{n}}=a+\left( n-1 \right)d$.

Complete step-by-step answer:

Here we have to insert six harmonic means between 3 and $\dfrac{6}{23}$. Also we have to find 22 times of the third term of the six terms. Let us consider six harmonic means to be ${{H}_{1}},{{H}_{2}},{{H}_{3}},{{H}_{4}},{{H}_{5}},{{H}_{6}}$. Then we can write the harmonic progression (H.P) as $3,{{H}_{1}},{{H}_{2}},{{H}_{3}},{{H}_{4}},{{H}_{5}},{{H}_{6}},\dfrac{6}{23}$.
Now, we know that if a series is in harmonic progression, then their reciprocal would be in arithmetic progression (A.P). Therefore, we get arithmetic progression (A.P) as, $\dfrac{1}{3},\dfrac{1}{{{H}_{1}}},\dfrac{1}{{{H}_{2}}},\dfrac{1}{{{H}_{3}}},\dfrac{1}{{{H}_{4}}},\dfrac{1}{{{H}_{5}}},\dfrac{1}{{{H}_{6}}},\dfrac{23}{6}$
Now, we have got an A.P with total 8 terms whose first term is $\dfrac{1}{3}$ and last term is $\dfrac{23}{6}$. We know that ${{n}^{th}}$ term of A.P is given by ${{a}_{n}}=a+\left( n-1 \right)d$, where a is the first term and d is the common difference of A.P. In this arithmetic series, as we have the first term, $a=\dfrac{1}{3}$. Therefore, we get ${{8}^{th}}$ term as ${{a}_{8}}=a+\left( 8-1 \right)d$
$\Rightarrow {{a}_{8}}=\dfrac{1}{3}+7d$
Since, we know the last term or the ${{8}^{th}}$ term of the given A.P is $\dfrac{23}{6}$, we get,
$\Rightarrow \dfrac{23}{6}=\dfrac{1}{3}+7d$
$\Rightarrow 7d=\dfrac{23}{6}-\dfrac{1}{3}$
$\Rightarrow 7d=\dfrac{23-2}{6}$
$\Rightarrow 7d=\dfrac{21}{6}$
$\Rightarrow d=\dfrac{21}{7\times 6}$
Therefore, we get $d=\dfrac{3}{6}=\dfrac{1}{2}$.
Hence, we get the second term of A.P $=\dfrac{1}{{{H}_{1}}}$ as ${{a}_{2}}=a+\left( 2-1 \right)d\Rightarrow a+d$,
$=\dfrac{1}{3}+\dfrac{1}{2}$
$=\dfrac{2+3}{6}$
Therefore, we get ${{a}_{2}}=\dfrac{1}{{{H}_{1}}}=\dfrac{5}{6}$, rearranging we get ${{H}_{1}}=\dfrac{6}{5}$.
Similarly, we get the third term of A.P $=\dfrac{1}{{{H}_{2}}}$ as ${{a}_{3}}=a+\left( 3-1 \right)d\Rightarrow a+2d$,
$=\dfrac{1}{3}+2.\left( \dfrac{1}{2} \right)$
${{a}_{3}}=\dfrac{4}{3}$
Therefore, we get ${{a}_{3}}=\dfrac{1}{{{H}_{2}}}=\dfrac{4}{3}$, rearranging, we get ${{H}_{2}}=\dfrac{3}{4}$.
Similarly, we get the fourth term of A.P $=\dfrac{1}{{{H}_{3}}}$ as ${{a}_{4}}=a+\left( 4-1 \right)d\Rightarrow a+3d$,
$=\dfrac{1}{3}+\dfrac{3}{2}$
$=\dfrac{2+9}{6}$
${{a}_{4}}=\dfrac{11}{6}$
Therefore, we get ${{a}_{4}}=\dfrac{1}{{{H}_{3}}}=\dfrac{11}{6}$, rearranging,
we get ${{H}_{3}}=\dfrac{6}{11}$.
Similarly, we get the fifth term of A.P $=\dfrac{1}{{{H}_{4}}}$ as ${{a}_{5}}=a+\left( 5-1 \right)d\Rightarrow a+4d$,
$=\dfrac{1}{3}+\dfrac{4}{2}$
$=\dfrac{1}{3}+2$
${{a}_{5}}=\dfrac{7}{3}$
Therefore, we get ${{a}_{5}}=\dfrac{1}{{{H}_{4}}}=\dfrac{7}{3}$, rearranging we get ${{H}_{4}}=\dfrac{3}{7}$.
Similarly, we get the sixth term of A.P $=\dfrac{1}{{{H}_{5}}}$ as ${{a}_{6}}=a+\left( 6-1 \right)d=a+5d$,
$=\dfrac{1}{3}+\dfrac{5}{2}$
$=\dfrac{2+15}{6}$
${{a}_{6}}=\dfrac{17}{6}$
Therefore, we get ${{a}_{6}}=\dfrac{1}{{{H}_{5}}}=\dfrac{17}{6}$, rearranging we get ${{H}_{5}}=\dfrac{6}{17}$.
Similarly, the seventh term of A.P $=\dfrac{1}{{{H}_{6}}}$ as ${{a}_{7}}=a+\left( 7-1 \right)d=a+6d$,
$=\dfrac{1}{3}+\dfrac{6}{2}$
$=\dfrac{1}{3}+3$
${{a}_{7}}=\dfrac{10}{3}$
Therefore, we get ${{a}_{7}}=\dfrac{1}{{{H}_{6}}}=\dfrac{10}{3}$, rearranging we get ${{H}_{6}}=\dfrac{3}{10}$.
Therefore, we get the 6 harmonic means between 3 and $\dfrac{6}{23}$ as: ${{H}_{1}}=\dfrac{6}{5}$, ${{H}_{2}}=\dfrac{3}{4}$, ${{H}_{3}}=\dfrac{6}{11}$, ${{H}_{4}}=\dfrac{3}{7}$, ${{H}_{5}}=\dfrac{6}{17}$ and ${{H}_{6}}=\dfrac{3}{10}$.
Now, we get 22 times of the third term of 6 terms $=22{{H}_{3}}$
$=22.\dfrac{6}{11}$
$=2\times 6\Rightarrow 12$
Therefore, 22 times of the third term of H.P is 12.

Note: Students should always use results of A.P in the questions of H.P because H.P is the reciprocal of A.P and we have the general formulas for A.P only. Also, students often forget to take the reciprocal of the term after finding the ${{n}^{th}}$ term of A.P to get H.P, so this mistake must be avoided.