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. = $12$.
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. = $5$.
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}$.
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. = $\dfrac{7y}{2} + 4$.
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