If the reprocals of 11 terms of harmonic sequence is 110 find the sixth term of harmonic sequence

Answer 1

In a harmonic sequence, the reciprocals of the terms form an arithmetic sequence. Let the terms of the harmonic sequence be represented as $a_1,a_2,a_3,\dots ,a_{11}$. Then the reciprocals can be represented as:

$$\frac{1}{a_1},\frac{1}{a_2},\frac{1}{a_3},\dots ,\frac{1}{a_{11}}$$

Let’s denote the first term of the reciprocal sequence (which is an arithmetic sequence) as $\frac{1}{a}$ and the common difference as $d$. Thus, the terms of the reciprocal sequence can be expressed as follows:

$$\frac{1}{a_1}=\frac{1}{a}\frac{1}{a_2}=\frac{1}{a}+d\frac{1}{a_3}=\frac{1}{a}+2d⋮\frac{1}{a_{11}}=\frac{1}{a}+10d$$

The sum of the reciprocals of 11 terms is given to be 110:

$$\sum _{n=0}^{10}(\frac{1}{a}+nd)=110$$

This expands to:

$$11\left(\frac{1}{a}\right)+d(0+1+2+\dots +10)=110$$

The sum of the first 10 integers is:

$$0+1+2+\dots +10=\frac{10\cdot 11}{2}=55$$

Putting it back into the equation gives us:

$$11\left(\frac{1}{a}\right)+55d=110$$

Subtracting 110 from both sides:

$$11\cdot \frac{1}{a}+55d=110$$

Now, we can simplify by dividing every term by 11:

$$\frac{1}{a}+5d=10$$

From here, we can express $\frac{1}{a}$ in terms of $d$:

$$\frac{1}{a}=10-5d$$

Next, to find the sixth term of the harmonic sequence, we need $a_6$:

$$\frac{1}{a_6}=\frac{1}{a}+5d=(10-5d)+5d=10$$

Thus, we have:

$$a_6=\frac{1}{10}$$

Therefore, the sixth term of the harmonic sequence is

$$\boxed{{\displaystyle \frac{1}{10}}}$$