Question 13, Exercise 4.2

Solutions of Question 13 of Exercise 4.2 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 A.M. between $7$ and $17$

Solution.

Here $a=7$ and $b=17$.
Now \begin{align*} \text{A.M.} &= \frac{a + b}{2}\\ &= \frac{7 + 17}{2} \\ &= \frac{24}{2} = 12. \end{align*} Hence A.M. = $12$.

Find A.M. between $3+3 \sqrt{2}$ and $7-3 \sqrt{2}$

Solution.

Here $a=3+3\sqrt{2}$ and $b=7-3\sqrt{2}$.
Now \begin{align*} \text{A.M.} &= \frac{a + b}{2}\\ &= \frac{(3 + 3 \sqrt{2}) + (7 - 3 \sqrt{2})}{2} \\ &= \frac{3 + 7 + 3 \sqrt{2} - 3 \sqrt{2}}{2} \\ &= \frac{10}{2}= 5. \end{align*} Hence A.M. = $5$.

Find A.M. between $7 \sqrt{5}$ and $\sqrt{5}$

Solution. Here $a=7\sqrt{5}$ and $b=\sqrt{5}$.
Now \begin{align*} \text{A.M.} &= \frac{a + b}{2}\\ &= \frac{7 \sqrt{5} + \sqrt{5}}{2} \\ &= \frac{(7 + 1) \sqrt{5}}{2} \\ &= \frac{8 \sqrt{5}}{2} = 4 \sqrt{5} \end{align*} Hence A.M. = $ 4 \sqrt{5}$.

Find A.M. between $2y+5$ and $5y+3$

Solution. Here $a=2y+5$ and $b=5y+3$.
Now \begin{align*} \text{A.M.} &= \frac{a + b}{2}\\ &= \frac{(2y + 5) + (5y + 3)}{2} \\ &= \frac{2y + 5y + 5 + 3}{2} \\ &= \frac{7y + 8}{2} \\ &= \frac{7y}{2} + \frac{8}{2} \\ &= \frac{7y}{2} + 4 \end{align*} Hence A.M. = $\dfrac{7y}{2} + 4$.

< Question 11 & 12 Question 14 & 15 >