Question

Calculate the $n^{th}$ term of the harmonic sequence $\frac{1}{3},\frac{1}{6},\frac{1}{9}\dots$ is

• A. $3^n$
• B. $\frac{3}{3^n}$
• C. $\frac{-1}{3^n}$
• D. $\frac{1}{3n}$

Answer 1

The correct option is D $\frac{1}{3n}$

$n^{th}$ term of a H.P. is given by $\frac{1}{a+(n-1)d}$

Where $a$ is the first term and $d$ is the common difference of H.P.

for the given series, we have

$a=3,d=3$

$n^{th}$ term will be

$\Rightarrow \frac{1}{3+3(n-1)}=\frac{1}{3n}$