====== Question 3 and 4, Exercise 4.2 ======
Solutions of Question 3 and 4 of Exercise 4.2 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 11th term of the arithmetic sequence $0.07,0.12,0.7, \ldots$

** Solution. **

Given
$$0.07,0.12,0.7, \ldots$$
From this, we have $a_1 = 0.07$, $d=0.05$, $a_{11}=?$.\\ 
Now
\begin{align*}
a_n&=a_1+(n-1)d \\
\implies a_{11}&= 0.07+(11-1)(0.05)\\
&=0.07+(10)(0.05)\\
&=0.57
\end{align*}
Hence $a_{11}=0.57.$ GOOD

=====Question 4=====
The third term of an arithmetic sequence is 14 and the ninth term is -1 . Find the first four terms of the sequence.

** Solution. **
Given: $a_3 = 14$ and  $a_9 = -1$.

The nth term of the arithmetic sequence is given as
$$a_n = a_1 + (n-1)d$$
Given
\begin{align*}
& a_3 = 14 \\
\implies & a_1 + 2d = 14 \quad \cdots (1)
\end{align*}
Also
\begin{align*}
& a_9 = -1 \\
\implies & a_1 + 8d = -1 \quad \cdots (2)
\end{align*}

Now, subtract equation (1) from equation (2):
\begin{align*}
\begin{array}{ccc}
a_1 & + 8d &= -1\\
\mathop{}\limits_{-}a_1 &\mathop+\limits_{-} 2d &= \mathop{}\limits_{-}14 \\ \hline
& 6d &= -15\\
\end{array}&\\
\implies \boxed{d = -\frac{5}{2}} \quad \\
\end{align*}

Putting the value $d$ in (1)
\begin{align*}
& a_1 + 2\left(-\frac{5}{2}\right) = 14 \\
\implies & a_1 = 14 + 5 \\
\implies & \boxed{a_1 = 19}
\end{align*}

Now,
\begin{align*}
a_2 &= a_1 + d = 19 - \frac{5}{2} = \frac{33}{2} \\
a_3 &= a_2 + d = \frac{33}{2} - \frac{5}{2} = 14 \\
a_4 &= a_3 + d = 14 - \frac{5}{2} = \frac{23}{2}
\end{align*}

Hence the first four terms are $19, \dfrac{33}{2}, 14, \dfrac{23}{2}$. GOOD
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