Question 9 and 10, Exercise 4.1
Solutions of Question 9 and 10 of Exercise 4.1 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 9
The $n$th term of the sequence is given, find the first 4 terms; the 10th term, $a_{10}$ and the 15 th term, $a_{15}$: $a_{n}=(-1)^{n}(n+3)$
Solution.
Do yourself.
Question 10
The $n$th term of the sequence is given, find the first 4 terms; the 10th term, $a_{10}$ and the 15 th term, $a_{15}$: $$a_{n}=(-1)^{n+1}(3 n-5).$$
Solution.
Given:
$$a_n = (-1)^{n+1}(3n - 5).$$
we can compute the following terms:
\begin{align*}
a_1 &= (-1)^{1+1}(3(1) - 5) = (1)(3 - 5) = -2 \\
a_2 &= (-1)^{2+1}(3(2) - 5) = (-1)(6 - 5) = -1 \\
a_3 &= (-1)^{3+1}(3(3) - 5) = (1)(9 - 5) = 4 \\
a_4 &= (-1)^{4+1}(3(4) - 5) = (-1)(12 - 5) = -7
\end{align*}
Now
\begin{align*}
a_{10} &= (-1)^{10+1}(3(10) - 5) = (-1)(30 - 5) = -25 \\
a_{15} &= (-1)^{15+1}(3(15) - 5) = (1)(45 - 5) = 40
\end{align*}
So, $a_1 = -2$, $a_2 = -1$, $a_3 = 4$, $a_4 = -7$, $a_{10} = -25$, $a_{15} = 40$. 
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