====== Question 1, Exercise 4.2 ======
Solutions of Question 1 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 1(i)=====
Find the first four terms of the arithmetic sequence with $a_{1}=4, d=3$

** Solution. **

Given: $a_1= 4$, $d=3$.\\
The general term of an arithmetic sequence is:
$$a_n = a_1 + (n - 1)d.$$
Now
\begin{align*}
a_2&=4+(2-1)3=4+3=7\\
a_3 &= 4+ (3-1) 3 = 4 + 6 = 10\\
a_4&=4+(4-1)3=4+9=13
\end{align*}
Hence $a_1=4$, $a_2=7$, $a_3=10$, $a_4=13$. GOOD

=====Question 1(ii)=====
Find the first four terms of the arithmetic sequence with $a_1=7$, $d=5$

** Solution. **

Given: $a_1= 7$, $d=5$.\\
The general term of an arithmetic sequence is:
$$a_n = a_1 + (n - 1)d.$$
Now
\begin{align*}
a_2&=7+(2-1)(5)=7+5=12\\
a_3 &= 7+ (3-1)(5) = 7 + 10 = 17\\
a_4&=7+(4-1)(5)=7+15=22
\end{align*}
Hence $a_1=7$, $a_2=12$, $a_3=17$, $a_4=22$. GOOD


=====Question 1(iii)=====
Find the first four terms of each arithmetic sequence. $a_{1}=16$, $d=-2$.

** Solution. **

Given: $a_1= 16$, $d=-2$.\\
We have
$$a_n = a_1 + (n - 1)d.$$
Now
\begin{align*}
a_2&=16+(2-1)(-2)=16-2=14\\
a_3 &= 16+ (3-1)(-2) = 16 - 4 = 12\\
a_4&=16+(4-1)(-2)=16-6=10
\end{align*}
Hence $a_1=16$, $a_2=14$, $a_3=12$, $a_4=10$. GOOD


=====Question 1(iv)=====
Find the first four terms of the arithmetic sequence. $a_1=38$, $d=-4$.

** Solution. **

Given: $a_1= 38$, $d=-4$.\\
We have
$$a_n = a_1 + (n - 1)d.$$
Now
\begin{align*}
a_2&=38+(2-1)(-4)=38-4=34\\
a_3 &= 38+ (3-1)(-4) = 38 - 8 = 30\\
a_4&=38+(4-1)(-4)=38-12=26
\end{align*}
Hence $a_1=38$, $a_2=34$, $a_3=30$, $a_4=26$. GOOD




=====Question 1(v)=====
Find the first four terms of each arithmetic sequence. $a_{1}=\frac{3}{4}, d=\frac{1}{4}$

** Solution. **

Given: $a_1=\frac{3}{4}$, $d=\frac{1}{4}$.\\
We have
$$a_n = a_1 + (n - 1)d.$$
Now
\begin{align*}
a_2&=\frac{3}{4}+(2-1)\cdot\frac{1}{4}=\frac{3}{4}+\frac{1}{4}=\frac{4}{4}=1\\
a_3 &= \frac{3}{4}+(3-1)\cdot\frac{1}{4}=\frac{3}{4}+\frac{2}{4}=\frac{5}{4}\\
a_4&=\frac{3}{4}+(4-1)\cdot\frac{1}{4}=\frac{3}{4}+\frac{3}{4}=\frac{6}{4}=1.5
\end{align*}
Hence $a_1=\frac{3}{4}$, $a_2=1$, $a_3=\frac{5}{4}$, $a_4=\frac{6}{4}=1.5$.


=====Question 1(vi)=====
Find the first four terms of each arithmetic sequence. $a_{1}=\frac{3}{8}, d=\frac{5}{8}$

** Solution. **

Given: $a_1=\frac{3}{8}$, $d=\frac{5}{8}$.\\
We have
$$a_n = a_1 + (n - 1)d.$$
Now
\begin{align*}
a_2 &= \frac{3}{8} + (2-1) \cdot \frac{5}{8} = \frac{3}{8} + \frac{5}{8} = \frac{8}{8} = 1\\
a_3 &= \frac{3}{8} + (3-1) \cdot \frac{5}{8} = \frac{3}{8} + 2 \cdot \frac{5}{8} = \frac{3}{8} + \frac{10}{8} = \frac{13}{8}\\
a_4 &= \frac{3}{8} + (4-1) \cdot \frac{5}{8} = \frac{3}{8} + 3 \cdot \frac{5}{8} = \frac{3}{8} + \frac{15}{8} = \frac{18}{8} = \frac{9}{4}
\end{align*}
Hence $a_1=\frac{3}{8}$, $a_2=1$, $a_3=\frac{13}{8}$, $a_4=\frac{9}{4}$.




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