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.

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, $a_1 = 1$, $a_2 = \frac{-1}{2}$, $a_3 = \frac{1}{4}$, $a_4 = \frac{-1}{8}$, $a_{10} = \frac{-1}{512}$, $a_{15} = \frac{1}{16384}$.

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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