====== Question 5 & 6, Exercise 4.6 ======
Solutions of Question 5 & 6 of Exercise 4.6 of Unit 04: Sequence and Series. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan. 

=====Question 5======
Find the indicated term of the harmonic progression. $\dfrac{1}{27}, \dfrac{1}{20}, \dfrac{1}{13}, \ldots \quad$ nth term.

** Solution. **

\begin{align*}
&\frac{1}{27}, \frac{1}{20}, \frac{1}{13}, \ldots \quad \text{ is in H.P.} \\
&27, 20, 13, \ldots \quad \text{ is in A.P.}
\end{align*}
Here $a_1 = 27$, $d = 20 - 27 = -7$, $a_n=?$
	
The general term of the A.P. is given as
$$
a_n = a_1 + (n-1)d.
$$
Thus,
\begin{align*}
a_n &= 27 + (n-1)(-7) \\
&= 27 - 7(n-1) \\
&= 27 - 7n + 7 \\
&= 34 - 7n.
\end{align*}
	
Hence, the $n$th term in H.P. is $\frac{1}{34 - 7n}.$

=====Question 6=====
Find the indicated term of the harmonic progression. $\frac{1}{2}, \frac{1}{2 \frac{1}{2}}, \frac{1}{3}, \frac{1}{3 \frac{1}{2}}, \ldots$ nth term.

** Solution. **


\begin{align*}
&\frac{1}{2}, \frac{1}{2 \frac{1}{2}}, \frac{1}{3}, \frac{1}{3 \frac{1}{2}}, \ldots \quad \text{ is in H.P.} \\
&2, 2 \frac{1}{2}, 3, 3 \frac{1}{2}, \ldots \quad \text{ is in A.P.}
\end{align*}
Here $a_1 = 2$, $d = 2 \frac{1}{2} - 2 = \frac{1}{2}$, $a_n=?$

The general term of the A.P. is given as
$$
a_n = a_1 + (n-1)d.
$$
Thus,
\begin{align*}
a_n &= 2 + (n-1)\left(\frac{1}{2}\right) \\
&= 2 + \frac{n-1}{2} \\
&= \frac{4}{2} + \frac{n-1}{2} \\
&= \frac{4 + n - 1}{2} \\
&= \frac{n+3}{2}.
\end{align*}
Hence, the $n$th term in H.P. is $\dfrac{1}{\frac{n+3}{2}}$ or $\dfrac{2}{n+3}$. m(






====Go to ====

<text align="left"><btn type="primary">[[math-11-nbf:sol:unit04:ex4-6-p2|< Question 3 & 4]]</btn></text>
<text align="right"><btn type="success">[[math-11-nbf:sol:unit04:ex4-6-p4|Question 7 & 8 >]]</btn></text>