====== Question 16 Exercise 4.2 ======
Solutions of Question 16 of Exercise 4.2 of Unit 04: Sequence and Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

=====Question 16=====
Insert five arithmetic means between $5$ and $8$ and show that their sum is five times the arithmetic mean between $5$ and $8$. GOOD
====Solution====
Let $A_1, A_2, A_3, A_4, A_5$ be five arithmetic means between $5$ and $8$. Then $5, A_1, A_2, A_3, A_4, A_5, 8$ are in A.P, where
$$a_1=5 \text{ and } a_7=8.$$
As we have
\begin{align}&a_7=a+6d\\
\implies &8=5+6d\\
\implies &6d=8-5\\
\implies &d=\dfrac{3}{6}=\dfrac{1}{2}.
\end{align}
Now
\begin{align}
A_1&=a+d=5+\dfrac{1}{2}=\dfrac{11}{2},\\
A_2&=a+2 d=5+2 \cdot \dfrac{1}{2}=6,\\
A_3&=a+3 d=5+3 \cdot \dfrac{1}{2}=\dfrac{13}{2},\\
A_4&=a+4 d=5+4 \cdot \dfrac{1}{2}=7,\\
A_5&=a+5 d=5+5 \cdot \dfrac{1}{2}=\dfrac{15}{2}.\end{align}
Hence $\dfrac{11}{2},6,\dfrac{13}{2},7, \dfrac{15}{2}$ are five A.Ms between $5$ & $8$.

Now \begin{align}A_1&+A_2+A_3+A_4+A_5 \\
&=\dfrac{11}{2}+6+\dfrac{13}{2}+7+\dfrac{15}{2}\\
&=\dfrac{11}{2}+\dfrac{12}{2}+\dfrac{13}{2}+\dfrac{14}{2}+\dfrac{15}{2}\\
&=\dfrac{11+12+13+14+15}{2}\\
&=\dfrac{65}{2}---(i)\end{align}
Let $A$ be arithmetic mean between 5 and 8. Then
$$A=\dfrac{a+b}{2}=\dfrac{5+8}{2}=\dfrac{13}{2}$$
Multiplying both side of the arithmetic mean by 5 , we get\\
$$5 A=\dfrac{65}{2}---(ii)$$
From (i) and (ii), we get\\
\begin{align}A_1+A_2+A_3+A_4+A_5=5A.\end{align}
Hence sum of five A.Ms between 5 and 8 is five times their A.M. GOOD


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