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.

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 $40$.

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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