====== Question 7 & 8, Exercise 4.6 ======
Solutions of Question 7 & 8 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 7=====
Find the 14 th term of H.P. $\frac{1}{4}, \frac{1}{7}, \frac{1}{10}, \frac{1}{13}, \ldots$

** Solution. **

Given the $ \frac{1}{4}, \frac{1}{7}, \frac{1}{10}, \frac{1}{13}, \ldots $ is an arithmetic sequence.
	
Here, $ a_1 = \frac{1}{4} $, $d = \frac{1}{7} - \frac{1}{4} = -\frac{3}{28},$
$ n = 14$
	
The general term of the A.P. is given as:
$$a_n = a_1 + (n-1)d.$$
Thus,
\begin{align*}
a_{14} &= \frac{1}{4} + (13)\left(-\frac{3}{28}\right) \\
&= \frac{1}{4} + 13\left(-\frac{3}{28}\right) \\
&= \frac{1}{4} - \frac{39}{28}\\
&= -\frac{8}{7} \text{ is in A.P.}
\end{align*}
Thus, the 14th term in the harmonic progression is: $-\frac{7}{8}.$ GOOD m(





=====Question 8=====
$7,4,1, \ldots$ is arithmetic sequence, find the 17th term in H.P.

** Solution. **

Given $7,4,1, \ldots$ is arithmetic sequence.

Here $a_1=7$, $d=4-7=-3$, $n=17$.

The general term of A.P is given as 
$$
a_n=a_1+(n-1)d.
$$
Thus
\begin{align*}
a_{17}&=7+(16)(-3) \\
&= 7-48 \\
&= -41 \text{ is in A.P.}
\end{align*}
Hence 17th term in H.P is $-\dfrac{1}{41}$. GOOD






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