Question

Find the $4^{th}$ term of an H.P, whose $7^{th}$ term is $\frac{1}{20}$ and ${13}^{th}$ term is $\frac{1}{38}$.

Answer 1

$\begin{array}{c}\frac{1}{a+\left(n-1\right)d}=T_n\left[\begin{array}{c}\because GeneralterminHP\\ \frac{1}{a+\left(n-1\right)d}\\ Whereafirsttermand\\ d=commondifference\end{array}\right]\\ T_7=\frac{1}{20}\left(given\right)\\ T_{13}=\frac{1}{38}\left(given\right)\\ T_7=\frac{1}{a+\left(7-1\right)d}\\ \frac{1}{20}=\frac{1}{a+6d}\\ a+6d=20\to \left(i\right)\\ T_{13}=\frac{1}{a+12d}\\ \frac{1}{38}=\frac{1}{a+12d}\\ a+12d=38\to \left(ii\right)\\ Substractequation\left(ii\right)from\left(i\right)\\ a+12d=38\\ -\underset{–}{a+6d=20}\\ 6d=18\\ d=3\\ Puttingd=3ine{q}^n\left(i\right)\\ a+18=20\\ a=2\\ Then,\\ T_4=\frac{1}{a+3d}=\frac{1}{2+3\times 3}=\frac{1}{11}Ans.\end{array}$