Question

A harmonic progression is a sequence of numbers such that their reciprocals are in arithmetic progression.
Let $S_n$ represent the sum of the first $n$ terms of the harmonic progression; for example, $S_3$ represents the sum of the first three terms. If the first three terms of a harmonic progression are $3,4,6$, then:

• A. $S_4=20$
• B. $S_4=25$
• C. $S_5=49$
• D. $S_6=49$
• E. $S_2=\frac{1}{2}{S}_4$

Answer 1

The correct option is B $S_4=25$

$H.P=3,4,6$

Now, let us take the arithmetic progression from the given H.P

$A.P=\frac{1}{3},\frac{1}{4},\frac{1}{6}$

Here, $T_2-T_1=T_3-T_2=\frac{-1}{12}=d$

So, in order to find the $4^{th}$ term of an $A.P$, use the formula,

The $n^{th}$ term of an $A.P=a+(n-1)d$

Here, $a=\frac{1}{3},$ $d=\frac{-1}{12}$

Now, we have to find the $4^{t}h$ term.

So, take $n=4$

Now put the values in the formula.

$4^{th}$ term of an $A.P=\frac{1}{3}+(4-1)\times \left(\frac{-1}{12}\right)$

$=\frac{1}{3}-\frac{1}{4}$

$=\frac{4-3}{12}$

$=\frac{1}{12}$

$4^{th}$ term of an $H.P=\frac{1}{4^{th}termofanA.P}=\frac{1}{1/12}=12$

Now,

$S_4=3+4+6+12=25$