# Question 16 and 17, Exercise 4.2

Solutions of Question 16 and 17 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 16

Find the two arithmetic means between $5$ and $17$.

** Solution. **
Let $A_1$ and $A_2$ be two arithmetic means between $5$ and $17$.

Then $5$, $A_1$, $A_2$, $17$ are in A.P.

Here $a_1=5$ and $a_4=17$.

Since nth term of A.P is given as
$$a_n=a_1+(n-1)d.$$
Thus
\begin{align*}
&a_4 = a_1 + 3d \\
\implies & 17=5+3d\\
\implies & 3d=12\\
\implies & \boxed{d=4}.\end{align*}
Now
\begin{align*}
A_1 &= a_2= a_1+d \\
&=5+4=9 \end{align*}
and
\begin{align*}
&A_2= a_3=a_1+2d\\
&= 5 + 2(4) \\
&=13 \end{align*}
Hence $A_1 = 9$ and $A_2 = 13$.

## Question 16

Find the two arithmetic means between $5$ and $17$.

** Solution. **

Let $A_1$, $A_2$ and $A_3$ be thre arithmetic means between $2$ and $-18$.

Then $2$, $A_1$, $A_2$, $A_3$, $-18$ are in A.P.

Here $a_1=2$ and $a_5=-18$.

Since nth term of A.P is given as
$$a_n=a_1+(n-1)d.$$
Thus
\begin{align*}
&a_5 = a_1 + 4d \\
\implies & -18=2+4d\\
\implies & 4d=-20\\
\implies & \boxed{d=-5}.\end{align*}
Now
\begin{align*}
A_1 &= a_2= a_1+d \\
&=2-5=-3 \end{align*}
\begin{align*}
&A_2= a_3=a_1+2d\\
&= 2 + 2(-5) \\
&=-8 \end{align*}
and
\begin{align*}
&A_3= a_4=a_1+3d\\
&= 2 + 3(-5) \\
&=-13 \end{align*}
Hence $A_1 = -3$, $A_2=-8$, A_3 = -13$.

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