Question 5, 6 and 7, Exercise 4.4

Solutions of Question 5, 6 and 7 of Exercise 4.4 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.

Find the first four terms of the geometric sequence.$a_{1}=3, r=-2$

Solution.

Given $a_{1}=3$ and $r=-2$. Use the formula $$a_{n}=a_{1} r^{n-1}.$$ Then \begin{align*} & a_{2}=a_{1} r=(3)(-2)= -6 \\ & a_{3}=a_{1} r^{2}=(3)(-2)^{2}=3 (4)= 12 \\ & a_{4}=a_{1} r^{3}=(3)(-2)^{3}=3 (-8) = -24 \end{align*} Hence $a_1=3$, $a_2=-6$, $a_3=12$, $a_4=-24$.

Find the first four terms of the geometric sequence. $a_{1}=27, r=-\frac{1}{3}$

Solution.

Given $a_{1}=27$ and $r=-\frac{1}{3}$. Use the formula $$a_{n}=a_{1} r^{n-1}.$$ Then \begin{align*} & a_{2}=a_{1} r=(27)\left(-\frac{1}{3}\right) = -9 \\ & a_{3}=a_{1} r^{2}=(27)\left(-\frac{1}{3}\right)^{2} = 27 \cdot \frac{1}{9} = 3 \\ & a_{4}=a_{1} r^{3}=(27)\left(-\frac{1}{3}\right)^{3} = 27 \cdot \left(-\frac{1}{27}\right) = -1 \end{align*} Hence $a_1=27$, $a_2=-9$, $a_3=3$, $a_4=-1$.

Find the first four terms of the geometric sequence. $\quad a_{1}=12, r=\frac{1}{2}$

Solution.

Given $a_{1}=12$ and $r=\frac{1}{2}$. Use the formula $$a_{n}=a_{1} r^{n-1}.$$ Then \begin{align*} & a_{2}=a_{1} r=(12)\left(\frac{1}{2}\right) = 6 \\ & a_{3}=a_{1} r^{2}=(12)\left(\frac{1}{2}\right)^{2} = 12 \cdot \frac{1}{4} = 3 \\ & a_{4}=a_{1} r^{3}=(12)\left(\frac{1}{2}\right)^{3} = 12 \cdot \frac{1}{8} = 1.5 \end{align*} Hence $a_1=12$, $a_2=6$, $a_3=3$, $a_4=1.5$.

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