====== 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 $A_2 = 13$. GOOD =====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$. GOOD ====Go to ==== [[math-11-nbf:sol:unit04:ex4-2-p9|< Question 14 & 15]]