Question 5 and 6, Exercise 4.2
Solutions of Question 5 and 6 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 5
Find an arithmetic sequence for $a_{17}=-40$ and $a_{28}=-73$, find $a_{1}$ and $d$. Write first five terms of the sequence.
Solution.
The nth term of the arithmetic sequence is given as
$$a_n=a_1+(n-1)d$$
Given
\begin{align*}
& a_{17} = -40 \\
\implies &a_1 + 16d = -40 \quad \cdots (1)
\end{align*}
Also
\begin{align*}
&a_{28}=-73\\
\implies &a_1 + 27d = -73 \quad \cdots (2)
\end{align*}
Now, subtract equation (1) from equation (2):
\begin{align*}
\begin{array}{ccc}
a_1& + 27d &= -73\\
\mathop{}\limits_{-}a_1 &\mathop+\limits_{-} 16d &= \mathop-\limits_{+}40 \\ \hline
& 11d &= -33\\
\end{array}&\\
\implies \boxed{d = -3} \quad \\
\end{align*}
Putting the value $d$ in (1)
\begin{align*}
& a_1 +16(-3) = -40 \\
\implies & a_1 = -40+48 \\
\implies & \boxed{a_1=8}
\end{align*}
Now,
\begin{align*}
a_2 &= a_1 + d = 8 - 3 = 5 \\
a_3 &= a_2 + d = 5 - 3 = 2 \\
a_4 &= a_3 + d = 2 - 3 = -1 \\
a_5 &= a_4 + d = -1 - 3 = -4
\end{align*}
Hence the first five terms are $8, 5, 2, -1, -4$ and $a_1 = 8$, $d=-3$. 
Question 6
The fifth term of an arithmetic sequence is 19 and 11 th term is 43. Find the first term and 87 th term.
Solution.
Given: $a_5 = 19$ and $a_{11} = 43$.
The nth term of the arithmetic sequence is given as $$a_n = a_1 + (n-1)d$$
Given \begin{align*} & a_5 = 19 \\ \implies & a_1 + 4d = 19 \quad \cdots (1) \end{align*} Also \begin{align*} & a_{11} = 43 \\ \implies & a_1 + 10d = 43 \quad \cdots (2) \end{align*}
Now, subtract equation (1) from equation (2): \begin{align*} \begin{array}{ccc} a_1 & + 10d &= 43\\ \mathop{}\limits_{-}a_1 &\mathop+\limits_{-} 4d &= \mathop-\limits_{+}19 \\ \hline & 6d &= 24\\ \end{array}&\\ \implies \boxed{d = 4} \quad \\ \end{align*}
Putting the value $d$ in (1) \begin{align*} & a_1 + 4(4) = 19 \\ \implies & a_1 = 19 - 16 \\ \implies & \boxed{a_1 = 3} \end{align*}
Now,
\begin{align*}
a_{87} &= a_1 + 86d \\
&= 3 + 86(4) = 347
\end{align*}
Hence the $a_1 = 3$ and $a_{87} = 347$.
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