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$If the 10^{th} term of H.P is 21 and 21^{st} term of H.P is 10 then 420^{th} term is$

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Solution

$10^{th} terms of H . P = \frac{1}{a + 9 d} = 21, t_{21} = \frac{1}{a + 20 d} = 10$
$\begin{array}{l} \to a + 9 d = \frac{1}{21} - - (1) \\ and a + 20 d = \frac{1}{10} - - (2) \\ (2) - (1) = 11 d = \frac{1}{10} - \frac{1}{21} = \frac{11}{210} \\ \Rightarrow d = \frac{1}{210} a n d a = \frac{1}{210} \\ T_{420} = \frac{1}{\frac{1}{210} + \frac{419}{210}} = \frac{1}{2} \end{array}$

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Harmonic Progression

Remember concepts with our Masterclasses.

If the 10th term of H.P is 21 and 21st term of H.P is 10 then 420th term is

10th terms of H.P=1a+9d=21,t21=1a+20d=10→a+9d=121−−1and a+20d=110−−22−1=11d=110−121=11210⇒d=1210and a=1210T420=11210+419210=12

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$If the 10^{th} term of H.P is 21 and 21^{st} term of H.P is 10 then 420^{th} term is$