If for the harmonic progression, $t_{7}=\frac{1}{10}, t_{12}=\frac{1}{25}$, then $t_{20}=$
If for the harmonic progression, $t_{7}=\frac{1}{10}, t_{12}=\frac{1}{25}$, then $t_{20}=$
$\frac{1}{48}$
49
$\frac{1}{49}$
48
Solution
First term of an $\mathrm{AP}=10$ and the $12^{\text {th }}$ term $=25$. Considering corresponding AP $a+6 d=10$ and $a+11 d=25 d=3, a=-8$
$\Rightarrow T_{20}=a+19 d=8+57=49$
Hence, the $20^{\text {th }}$ term of the corresponding HP is $1 / 49$.