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Question: If \({{3}^{rd}}\) term of an H.P. is 7 and \[{{7}^{th}}\] term of H.P. is 3 then \[{{10}^{th}}\] ter...

If 3rd{{3}^{rd}} term of an H.P. is 7 and 7th{{7}^{th}} term of H.P. is 3 then 10th{{10}^{th}} term is?
a) 37\dfrac{3}{7}
b) 2110\dfrac{21}{10}
c) 107\dfrac{10}{7}
d) 37\dfrac{3}{7}

Explanation

Solution

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:
1a,1a+d,1a+2d,1a+3d,.....\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 nth{{n}^{th}} term of HP:
an=1a+(n−1)d{{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 10th{{10}^{th}} term.

Complete step by step answer:
Since it is given in the question that 3rd{{3}^{rd}} term of an H.P. is 7 and 7th{{7}^{th}} term of H.P. is 3.
So, by using the general formula for HP we can write:
a3=1a+2d=7......(1) a7=1a+6d=3......(2) \begin{aligned} & {{a}_{3}}=\dfrac{1}{a+2d}=7......(1) \\\ & {{a}_{7}}=\dfrac{1}{a+6d}=3......(2) \\\ \end{aligned}
We can rewrite equation (1) and (2) as:
a+2d=17......(3) a+6d=13......(4) \begin{aligned} & a+2d=\dfrac{1}{7}......(3) \\\ & a+6d=\dfrac{1}{3}......(4) \\\ \end{aligned}
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:
4d=13−17 4d=421 d=121......(5)\begin{aligned} & 4d=\dfrac{1}{3}-\dfrac{1}{7} \\\ & 4d=\dfrac{4}{21} \\\ & d=\dfrac{1}{21}......(5) \end{aligned}
Substitute the value of d in equation (3), we get:
a+2(121)=17 a=17−221 a=121......(6)\begin{aligned} & a+2\left( \dfrac{1}{21} \right)=\dfrac{1}{7} \\\ & a=\dfrac{1}{7}-\dfrac{2}{21} \\\ & a=\dfrac{1}{21}......(6) \end{aligned}
Now, using the general formula, find the 10th{{10}^{th}} term of HP.
a10=1a+9d =1121+9×121 =2110\begin{aligned} & {{a}_{10}}=\dfrac{1}{a+9d} \\\ & =\dfrac{1}{\dfrac{1}{21}+9\times \dfrac{1}{21}} \\\ & =\dfrac{21}{10} \end{aligned}

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 nth{{n}^{th}} term of harmonic progression, it is reciprocal of nth{{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.