Question 7 and 8, Exercise 4.1
Solutions of Question 7 and 8 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 7
The $n$th term of the sequence is given, find the first 4 terms; the 10th term, $a_{10}$ and the 15 th term: $a_{15}$.$a_{n}=\left(\frac{-1}{2}\right)^{n-1}$
Solution.
Given:
$$a_n = \left( \frac{-1}{2} \right)^{n-1}.$$
Now
\begin{align*}a_1 &= \left( \frac{-1}{2} \right)^{1-1} = \left( \frac{-1}{2} \right)^0 = 1 \\
a_2 &= \left( \frac{-1}{2} \right)^{2-1} = \left( \frac{-1}{2} \right)^1 = \frac{-1}{2} \\
a_3 &= \left( \frac{-1}{2} \right)^{3-1} = \left( \frac{-1}{2} \right)^2 = \frac{1}{4} \\
a_4 &= \left( \frac{-1}{2} \right)^{4-1} = \left( \frac{-1}{2} \right)^3 = \frac{-1}{8}
\end{align*}
Now
\begin{align*}
a_{10} &= \left( \frac{-1}{2} \right)^{10-1} = \left( \frac{-1}{2} \right)^9 = \frac{-1}{512} \\
a_{15} &= \left( \frac{-1}{2} \right)^{15-1} = \left( \frac{-1}{2} \right)^{14} = \frac{1}{16384}
\end{align*}
So, $a_1 = 1$, $a_2 = \frac{-1}{2}$, $a_3 = \frac{1}{4}$, $a_4 = \frac{-1}{8}$, $a_{10} = \frac{-1}{512}$, $a_{15} = \frac{1}{16384}$. 
Question 8
The $n$th term of the sequence is given, find the first 4 terms; the 10th term, $a_{10}$ and the 15 th term: $a_{n}=(-1)^{2} \cdot n^{2}$
Solution.
Do yourself.
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