Question

If ${{3}^{rd}}$ term of an H.P. is 7 and \[{{7}^{th}}\] term of H.P. is 3 then \[{{10}^{th}}\] term is?
a) $\dfrac{3}{7}$
b) $\dfrac{21}{10}$
c) $\dfrac{10}{7}$
d) $\dfrac{3}{7}$

Answer 1

Hint: A harmonic progression or HP is a progression formed by taking the reciprocals of an arithmetic progression. So, the terms of HP can be given as:
$\dfrac{1}{a},\dfrac{1}{a+d},\dfrac{1}{a+2d},\dfrac{1}{a+3d},.....$where a is the first term and d is the common difference.
General formula for ${{n}^{th}}$ term of HP:
${{a}_{n}}=\dfrac{1}{a+\left( n-1 \right)d}$
Use the given formula and try to find the value of a and d from the given terms in the question and substitute to find the \[{{10}^{th}}\] term.

Complete step by step answer:
Since it is given in the question that ${{3}^{rd}}$ term of an H.P. is 7 and \[{{7}^{th}}\] term of H.P. is 3.
So, by using the general formula for HP we can write:
$\begin{align}
  & {{a}_{3}}=\dfrac{1}{a+2d}=7......(1) \\
 & {{a}_{7}}=\dfrac{1}{a+6d}=3......(2) \\
\end{align}$
We can rewrite equation (1) and (2) as:
$\begin{align}
  & a+2d=\dfrac{1}{7}......(3) \\
 & a+6d=\dfrac{1}{3}......(4) \\
\end{align}$
So, we have two equations in two variables. Solve the above equations to get the value of a and d.
Subtract equation (3) from equation (4), we get:
$\begin{align}
  & 4d=\dfrac{1}{3}-\dfrac{1}{7} \\
 & 4d=\dfrac{4}{21} \\
 & d=\dfrac{1}{21}......(5)
\end{align}$
Substitute the value of d in equation (3), we get:
$\begin{align}
  & a+2\left( \dfrac{1}{21} \right)=\dfrac{1}{7} \\
 & a=\dfrac{1}{7}-\dfrac{2}{21} \\
 & a=\dfrac{1}{21}......(6)
\end{align}$
Now, using the general formula, find the \[{{10}^{th}}\] term of HP.
$\begin{align}
  & {{a}_{10}}=\dfrac{1}{a+9d} \\
 & =\dfrac{1}{\dfrac{1}{21}+9\times \dfrac{1}{21}} \\
 & =\dfrac{21}{10}
\end{align}$

So, the correct answer is “Option B”.

Note: While dealing with harmonic progressions, always remember that harmonic progressions are reciprocal of arithmetic progression but not arithmetic progression itself. So, when we apply the general formula for ${{n}^{th}}$ term of harmonic progression, it is reciprocal of ${{n}^{th}}$ term of an arithmetic progression.
So, another way to solve the harmonic progression is to take the reciprocal of the given series and treat it as an arithmetic series.