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.

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 $A_2 = 13$.

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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