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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