====== 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. 


=====Question 13(i)=====
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. = $12$. GOOD

=====Question 13(ii)=====
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. = $5$. GOOD

=====Question 13(iii)=====
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}$. GOOD

=====Question 13(iv)=====
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. = $\dfrac{7y}{2} + 4$. GOOD

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