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- Question 14 Exercise 4.2
- Question 14(i)===== Insert three arithmetic means between 6 and 41. GOOD ====Solution==== Let $A_1, A_2, A_3$ be three arithmetic means between 6 and 41. Then $6, A_1, A_2, A_3, 41$ are in A.P.... ac{1}{4}.\end{align} Hence three arithmetic means between 6 and 41 are $$14\dfrac{3}{4},23\dfrac{1}{2},32\d... Question 14(ii)===== Insert four arithmetic means between 17 and 32. GOOD ====Solution==== Let $A_1, A_2, A
- Question 16 Exercise 4.2
- ====Question 16===== Insert five arithmetic means between $5$ and $8$ and show that their sum is five times the arithmetic mean between $5$ and $8$. GOOD ====Solution==== Let $A_1, A_2, A_3, A_4, A_5$ be five arithmetic means between $5$ and $8$. Then $5, A_1, A_2, A_3, A_4, A_5, 8$... },6,\dfrac{13}{2},7, \dfrac{15}{2}$ are five A.Ms between $5$ & $8$. Now \begin{align}A_1&+A_2+A_3+A_4+A_5
- Question 12 & 13 Exercise 4.2
- =====Question 13(i)===== Find the arithmetic mean between $12$ and $18$. GOOD ====Solution==== Here $a=12, ... }\\&=\dfrac{30}{2}=15.\end{align} Hence 15 is A.M between 12 and 18. GOOD =====Question 13(ii)===== Find the arithmetic mean between $\dfrac{1}{3}$ and $\dfrac{1}{4}$. ====Solution==... ===Question 13(iii)===== Find the arithmetic mean between $-6,-216$. GOOD ====Solution==== Here $a=-6, b=-2
- Question 9 Exercise 4.4
- ===Question 9(i)===== Insert five geometric means between $3 \dfrac{5}{9}=\dfrac{32}{9}\quad$ and $\quad40 ... , G_3, G_4$ and $G_5$ be the five geometric means between $\dfrac{32}{9}$ and $\dfrac{81}{2}$,\\ then $\dfr... ===Question 9(ii)===== Insert $6$ geometric means between $14$ and $-\dfrac{7}{64}$. ====Solution==== Let $... 3, G_4, G_5$ and $G_6$ be the six geometric means between $14$ and $-\dfrac{7}{64}$,\\ then $14, G_1, G_2,
- Question 10 Exercise 4.4
- estion 10===== Find two numbers if the difference between them is $48$ and their A.M exceeds their G.M by $... be $a$ and $b$ \\ Condition-$1$\\ The difference between them is $48$\\ Therefore, $$\quad a-b=48....(i)$$. The geometric mean between $a$ and $b$ is $$G=\sqrt{a b}$$ The arithmetic mean between $a$ and $b$ is $$A=\dfrac{a+b}{2}$$ Condition-$2$
- Question 9 & 10 Exercise 4.3
- estion 9===== Find the sum 'of all multiples of 9 between 300 and 700. ====Solution==== All the multiples of 9 between 300 and 700 are:\\ $$306,315,324,333, \ldots, 693... d{align} Hence, sum of all multiples of $9$ lying between $300$ and $700$ is equal to $21,978$. =====Quest
- Question 11 Exercise 4.4
- that the prodect of $\mathrm{n}$ geometric means between $a$ and $b$ is equal to the $nth$ power for the single geometric mean between them. ====Solution==== Let $G_1, G_2, G_9, \ldots, G_n$ be the $n$ geometric means between $a$ and $b$,\\ then $a, G_1, G_2, G_3, \ldots, G_
- Question 15 Exercise 4.2
- a^{n+1}+b^{n+1}}{a^n+b^n}$ is the arithmetic mean between $a$ and $b$. Where $a$ and $b$ are not zero simul... on==== Suppose $A$ represents the arithmetic mean between $a$ and $b$, then $$ A=\dfrac{a+b}{2}. --- (1) $$
- Question 17 Exercise 4.2
- ==Question 17===== There are $n$ arithmetic means between 5 and 32 such that the ratio of the 3rd and 7th m... 1, A_2, A_3, \ldots, A_n$ be $n$ arithmetic means between 5 and 32. Then $5, A_1, A_2, A_3, \ldots, A_n, 32
- Question 13 & 14 Exercise 4.3
- ==Question 14===== Insert enough arithmetic means between $1$ and $50$ so that the sum of the resulting ser... \\ so therefore there would need to be $16$ terms between $1$ and $50$. ====Go To==== <text align="left"><
- Question 12 Exercise 4.4
- frac{a^{n+1}+b^{n+1}}{a^n+b^n}$ is geometric mean between $a$ and $b$,\\ where $a$ and $b$ are not zero si... frac{a^{n+1}+b^{n-1}}{a^n+b^n}$ be geometric mean between $a$ and $b$, then\\ \begin{align}\dfrac{a^{n+1}+b