Question

The ${20}^{th}$ term from the last of sequence $\frac{1}{5},\frac{1}{10},\frac{1}{15},\frac{1}{20}.....\frac{1}{150}$ will be

• A. $\frac{1}{55}$
• B. $\frac{1}{245}$
• C. $\frac{1}{345}$
• D. $\frac{1}{445}$

Answer 1

Answer: (A)

$\frac{1}{55}$

Corresponding AP will be $5,10,15,20,....,150$

no. of terms is $n$ then

$\Rightarrow 150=5+(n-1)\times 5$

$\Rightarrow n=\frac{145}{5}+1$

$\Rightarrow n=30$

${20}^{th}$ term from the last means, $(30-20)+1={11}^{th}$ term from begining

$\Rightarrow {11}^{th}$ term of A.P. $=5+(11-1)\times 5=55$

$\therefore {11}^{th}$ term of HP is$=\frac{1}{55}$