# Question 22 and 23, Exercise 4.4

Solutions of Question 22 and 23 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.

## Question 22

Find the missing geometric means. $$8 , \_\_\_, \_\_\_, \_\_\_, \_\_\_, \dfrac{1}{4}$$

** Solution. **

We have $a_1=8$ and $a_6=\frac{1}{4}$.

Assume $r$ to be the common ratio. Then, by the general formula for the $n$th term, we have

$a_n = a_1 r^{n-1}.$

This gives

\begin{align*}
a_6 &= a_1 r^5 \\
\implies \frac{1}{4} &= 8 \cdot r^5 \\
\implies r^5 &= \frac{1}{4 \cdot 8} \\
\implies r^5 &= \frac{1}{32} \\
\implies r &= \left(\frac{1}{32}\right)^{\frac{1}{5}} \\
\implies r &= \frac{1}{2}.
\end{align*}
Thus, we can find the missing terms:
\begin{align*}
a_2 &= a_1 r = 8 \cdot \frac{1}{2} = 4, \\
a_3 &= a_1 r^2 = 8 \cdot \left(\frac{1}{2}\right)^2 = 8 \cdot \frac{1}{4} = 2, \\
a_4 &= a_1 r^3 = 8 \cdot \left(\frac{1}{2}\right)^3 = 8 \cdot \frac{1}{8} = 1, \\
a_5 &= a_1 r^4 = 8 \cdot \left(\frac{1}{2}\right)^4 = 8 \cdot \frac{1}{16} = \frac{1}{2}.
\end{align*}
Hence, the missing geometric means are $4$, $2$, $1$, and $d\frac{1}{2}$.

## Question 23

Find the missing geometric means. $$3 , \_\_\_ , 75$$

** Solution. **

We have $a_1=3$ and $a_3=75$. Assume $r$ to be the common ratio. Then, by the general formula for the $n$th term, we have $a_n = a_1 r^{n-1}.$ This gives \begin{align*} a_3 &= a_1 r^2 \\ \implies 75 &= 3 \cdot r^2 \\ \implies r^2 &= \frac{75}{3} \\ \implies r^2 &= 25 \\ \implies r &= 5. \end{align*} Thus, we can find the missing terms: \begin{align*} a_2 &= a_1 r = 3 \cdot 5 = 15. \end{align*} Hence, the missing geometric mean is $15$.

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