====== Question 3 & 4, Exercise 4.6 ======
Solutions of Question 3 & 4 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 3=====
Find the indicated term of the harmonic progression. $\frac{1}{18}, \frac{1}{13}, \frac{1}{8}, \ldots \quad 20$ th term.

** Solution. **

\begin{align*}
&\frac{1}{18}, \frac{1}{13}, \frac{1}{8}, \ldots \quad \text{ is in H.P.} \\
&18, 13, 8, \ldots \quad \text{ is in A.P.}
\end{align*}
Here $a_1 = 18$, $d = 13 - 18 = -5$ $a_{20}.$
	
The general term of the A.P. is given as
$$
a_n = a_1 + (n-1)d.
$$
Thus,
\begin{align*}
a_{20} &= 18 + (20-1)(-5) \\
&= 18 + 19(-5) \\
&= 18 - 95 \\
&= -77.
\end{align*}
	
Hence, the 20th term in H.P. is $-\frac{1}{77}.$



=====Question 4=====
Find the indicated term of the harmonic progression. $\frac{1}{4}, \frac{1}{9}, \frac{1}{14}, \ldots \quad$ nth term.

** Solution. **

\begin{align*}
&\frac{1}{4}, \frac{1}{9}, \frac{1}{14}, \ldots \quad \text{ is in H.P.} \\
&4, 9, 14, \ldots \quad \text{ is in A.P.}
\end{align*}
Here $a_1 = 4$, $d = 9 - 4 = 5$ $a_n=?$
	
The general term of the A.P. is given as
$$
a_n = a_1 + (n-1)d.
$$
Thus,
\begin{align*}
a_n &= 4 + (n-1)(5) \\
&= 4 + 5(n-1) \\
&= 4 + 5n - 5 \\
&= 5n - 1.
\end{align*}
	
Hence, the $n$th term in H.P. is $\frac{1}{5n - 1}.$




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