====== Question 24 and 25, Exercise 4.4 ======
Solutions of Question 24 and 25 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 24=====
Find the missing geometric means. 
$$5 , \_\_\_, \_\_\_, \_\_\_, 80$$

** Solution. **

We have $a_1=5$ and $a_5=80$.  
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_5 &= a_1 r^4 \\
\implies 80 &= 5 \cdot r^4 \\
\implies r^4 &= \frac{80}{5} \\
\implies r^4 &= 16 \\
\implies r &= 2.
\end{align*}
Thus, we can find the missing terms:
\begin{align*}
a_2 &= a_1 r = 5 \cdot 2 = 10, \\
a_3 &= a_1 r^2 = 5 \cdot 2^2 = 5 \cdot 4 = 20, \\
a_4 &= a_1 r^3 = 5 \cdot 2^3 = 5 \cdot 8 = 40.
\end{align*}
Hence, the missing geometric means are $10$, $20$, and $40$.

=====Question 25=====
Find the missing geometric means.
$$7 ,\_\_\_, \_\_\_, \_\_\_, 112$$

** Solution. **

We have $a_1=7$ and $a_6=112$.\\  
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 112 &= 7 \cdot r^5 \\
\implies r^5 &= \frac{112}{7} \\
\implies r^5 &= 16 \\
\implies r &= 2.
\end{align*}
Thus, we can find the missing terms:
\begin{align*}
a_2 &= a_1 r = 7 \cdot 2 = 14, \\
a_3 &= a_1 r^2 = 7 \cdot 2^2 = 7 \cdot 4 = 28, \\
a_4 &= a_1 r^3 = 7 \cdot 2^3 = 7 \cdot 8 = 56, \\
a_5 &= a_1 r^4 = 7 \cdot 2^4 = 7 \cdot 16 = 112.
\end{align*}
Hence, the missing geometric means are $14$, $28$, and $56.$



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