Harmonic progression

According to the question we have; $\dfrac 1{a+9d} = 21; \quad \dfrac 1{a+20d} = 10$ Simplifying them gives; $\dfrac 1{a+9d} = 21\implies 21a+189d = 1\\\dfrac 1{a+20d} = 10\implies 10a+200d = 1$ Therefore we can equate the left hand sides, $21a+189d = 10a+200d \\11a=11d \implies \boxed{a=d}\\\therefore10a+200d = 210a = 1\\\therefore a = d = \dfrac 1{210}$

Hence, 210th term is; $\dfrac 1{a+209d}=\dfrac 1{210d} =\dfrac 1{210\times\dfrac1{210}} = \boxed 1$

Since harmonic progression is the reciprocal of arithmetic progression, convert all the terms to arithmetic progression then compute for the $210^{th}$ term. Then take the reciprocal. So in arithmetic progression, $a_{10}=\dfrac{1}{21}$ and $a_{21}=\dfrac{1}{10}$ . The $n^{th}$ term of the AP is given by $a_n=a_m+(m-n)d$ .

Substituting, we have

$\dfrac{1}{10}=\dfrac{1}{21}+(21-10)d$ $\implies$ $\dfrac{1}{10}-\dfrac{1}{21}=11d$ $\implies$ $d=\dfrac{1}{210}$

Computing for the $210^{th}$ term of the AP , we have

$a_{210}=\dfrac{1}{10}+(210-21)\left(\dfrac{1}{210}\right)=\dfrac{1}{10}+\dfrac{189}{210}=1$

Since the reciprocal of $1$ is $1$ , the $210^{th}$ term of the HP is also $\boxed{1}$ .

a + (10-1)d = 1/21 which is 10th term |||| a + (21-1)d = 1/10 which is 21st term |||| Find values of a and d from above equations |||||| and put in equation |||||
a + (210-1)d which is 210th term.