Question 18 and 19, Exercise 4.4

Solutions of Question 18 and 19 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 nth term of the geometric sequence if $a_{1}=32, n=6, r=-\frac{1}{2}$

Solution.

Given $a_{1}=32$, $n=6$, $r=-\frac{1}{2}$.
General term of the geometric series is given by $a_{n}=a_{1} r^{n-1}.$ Thus $$\begin{align*} a_6 &= 32 \times \left(-\frac{1}{2}\right)^{6-1} \\ &= 32 \times \left(-\frac{1}{2}\right)^{5} \\ &= 32 \times \left(-\frac{1}{32}\right) \\ &= -1. \end{align*}$$ Hence $a_6=-1$.

Find the nth term of the geometric sequence if $a_{1}=16, n=8, r=\frac{1}{2}$

Solution.

Given $a_{1}=16$, $n=8$, $r=\frac{1}{2}$.
General term of the geometric series is given by $a_{n}=a_{1} r^{n-1}.$ Thus $$\begin{align*} a_8 &= 16 \times \left(\frac{1}{2}\right)^{8-1} \\ &= 16 \times \left(\frac{1}{2}\right)^{7} \\ &= 16 \times \frac{1}{128} \\ &= \frac{16}{128} = \frac{1}{8}. \end{align*}$$ Hence $a_8=\frac{1}{8}$.

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