Question 20 and 21, Exercise 4.4

Solutions of Question 20 and 21 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. $$3 , \_\_\_ , \_\_\_ , \_\_\_ , 48$$

Solution.

We have given $a_1=3$ and $a_5=48$.

Assume $r$ be common difference, then by general formula for nth term, we have $$a_n=ar^{n-1}.$$ This gives $$\begin{align*} &a_5=a_1 r^4 \\ \implies & 48=3r^4 \\ \implies & r^4 = 16 \\ \implies & r^4 = 2^4 \\ \implies & r = 2. \end{align*}$$ Thus $$\begin{align*} & a_2=a_1 r= (3)(2) = 6 \\ & a_3=a_1 r^2 = (3)(2)^2 = 12 \\ & a_4=a_1 r^3= (3)(2)^3=24. \end{align*}$$ Hence $6$, $12$, $24$ are required geometric means.

Find the missing geometric means. $$1 ,\_\_\_,\_\_\_, 8$$

Solution.

We have $a_1=1$ and $a_4=8$. 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 $$\begin{align*} a_4 &= a_1 r^3 \\ \implies 8 &= 1 \cdot r^3 \\ \implies r^3 &= 8 \\ \implies r^3 &= 2^3 \\ \implies r &= 2. \end{align*}$$ Thus, we can find the missing terms: $$\begin{align*} a_2 &= a_1 r = 1 \cdot 2 = 2, \\ a_3 &= a_1 r^2 = 1 \cdot 2^2 = 4. \end{align*}$$ Hence, the missing geometric means are $2$ and $4$.

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