Question 1: Find the $\text{A.M}$. between:

(i) $7$ and  $13$        (ii) $12$ and $-8$        (iii)  $(x-y)$  and $(x + y)$

i)       Let $A$ be the Arithmetic Mean of  $7$ and $13$.

Therefore $7, A, 13$ are in A.P.

$\Rightarrow A - 7 = 13 - A$

$\Rightarrow A =$ $\frac{7+13}{2}$ $= 10$

Hence the Arithmetic Mean of $7$ and $13$ is $10$.

ii)      Let $A$ be the Arithmetic Mean of  $12$ and $-8$.

Therefore $12, A, -8$ are in A.P.

$\Rightarrow A - 12 = -8 - A$

$\Rightarrow A =$ $\frac{12-8}{2}$ $= 2$

Hence the Arithmetic Mean of $12$ and $-8$ is $2$.

iii)     Let $A$ be the Arithmetic Mean of  $(x-y)$ and $(x+y)$.

Therefore $(x-y), A, (x+y)$ are in A.P.

$\Rightarrow A - (x-y) = (x+y) - A$

$\Rightarrow A =$ $\frac{x+y+x-y}{2}$ $= x$

Hence the Arithmetic Mean of $(x-y)$ and $(x+y)$ is $x$.

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Question 2: Insert $4$ $\text{A.M.s}$ between $4$ and $19$.

Let $A_1, A_2, A_3, A_4,$  be the 4 A.M.s between $4$ and $19$.

Therefore $4, A_1, A_2, A_3, A_4, 19$ are in A.P.

$a = 4 \ \ \ \ \ a_6 = 19$

We know $a_n = a + ( n-1) d$

$\therefore 19 = 4 + ( 6-1)(d) \Rightarrow 5d = 15 \Rightarrow d = 3$

Therefore

$A_1 = a + d = 4 + 3 = 7$

$A_2 = A_1 + d = 7 +3 = 10$

$A_3 = A_2 + d = 10 + 3 = 13$

$A_4 = A_3 + d = 13 + 3 = 16$

Therefore the 4 A.M.s between $4$ and $19$ are $7, 10, 13 ,16$

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Question 3: Insert $7$ $\text{A.M.s}$ between $2$ and  $17$.

Let $A_1, A_2, A_3, A_4, A_5, A_6, A_7$  be the 7 A.M.s between $2$ and $17$.

Therefore $2, A_1, A_2, A_3, A_4, A_5, A_6, A_7, 17$ are in A.P.

$a = 2 \ \ \ \ \ a_9 = 17$

We know $a_n = a + ( n-1) d$

$\therefore 17 = 2 + ( 9-1)(d) \Rightarrow 15 = 8d \Rightarrow d = \frac{15}{8}$

Therefore

$A_1 = a + d = 2 + \frac{15}{8} = \frac{31}{8}$

$A_2 = A_1 + d = \frac{31}{8} +\frac{15}{8} = \frac{46}{8}$

$A_3 = A_2 + d = \frac{46}{8} + \frac{15}{8} = \frac{61}{8}$

$A_4 = A_3 + d = \frac{61}{8} + \frac{15}{8} = \frac{76}{8}$

$A_5 = A_4 + d = \frac{76}{8} +\frac{15}{8} = \frac{91}{8}$

$A_6 = A_5 + d = \frac{91}{8} + \frac{15}{8} = \frac{106}{8}$

$A_7 = A_6 + d = \frac{106}{8} + \frac{15}{8} = \frac{121}{8}$

Therefore the 4 A.M.s between $4$ and $19$ are $\frac{31}{8},\frac{46}{8},\frac{61}{8},\frac{76}{8},\frac{91}{8},\frac{106}{8},\frac{121}{8}$

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Question 4: Insert $6$ $\text{A.M.s}$ between $15$ and $-13$

Let $A_1, A_2, A_3, A_4, A_5, A_6$  be the 6 A.M.s between $15$ and $-13$.

Therefore $15, A_1, A_2, A_3, A_4, A_5, A_6, -13$ are in A.P.

$a = 15 \ \ \ \ \ a_8 = -13$

We know $a_n = a + ( n-1) d$

$\therefore -13 = 15 + ( 8-1)(d) \Rightarrow 7d = -28 \Rightarrow d = -4$

Therefore

$A_1 = a + d = 15 -4 = 11$

$A_2 = A_1 + d = 11 -4 = 7$

$A_3 = A_2 + d = 7 -4 = 3$

$A_4 = A_3 + d = 3 -4 = -1$

$A_5 = A_4 + d = -1 -4 = -5$

$A_6 = A_5 + d = -5 -4 = 9$

Therefore the 6 A.M.s between $15$ and $-13$ are $11,7,3,-1,-5,-9$

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Question 5: There are $n$ $\text{A.M.s}$ between $3$ and $17$. The ratio of the last mean to the first mean is $3 : 1$. Find the value of $n$.

Let $A_1, A_2, \dots , A_n$ be the arithmetic means between $3$ and $17$

$\therefore a = 3$  and $a_{n+2} = 17$

We know $a_n = a + ( n-1) d$

$\Rightarrow 17 = 3 + ( n+2-1) d$

$\Rightarrow 14 = ( n+1) d$

$\Rightarrow d =$ $\frac{14}{n+1}$

$A_1 a + d = a +$ $\frac{14}{n+1}$ $=$ $\frac{3n+17}{n+1}$

$A_n = 3 + nd = 3 + n \Big($ $\frac{14}{n+1}$ $\Big) =$ $\frac{17n+3}{n+1}$

$\therefore$ $\frac{A_n}{A_1}$ $=$ $\frac{3}{1}$

$\Rightarrow$ $\frac{\frac{17n+3}{n+1}}{\frac{3n+17}{n+1}}$ $=$ $\frac{3}{7}$

$\Rightarrow$ $\frac{17n+3}{3n+17}$ $=$ $\frac{3}{1}$

$\Rightarrow 17n + 3 = 9n + 51$

$\Rightarrow 8n = 48$

$\Rightarrow n = 6$

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Question 6: Insert $\text{A.M.s}$, between $7$ and $71$ in such a way that $5^{th}$  $\text{A.M.}$ is $27$. Find the number of $\text{A.M.s}$.

Let $A_1, A_2, \dots , A_n$ be the arithmetic means between $7$ and $71$

Given $a = 7 \ \ \ \ \ a_6 = 27$

$\therefore a + 5d = 27$

$\Rightarrow 5d = 27 - 7$

$\Rightarrow d = 4$

$A_{n+2} = 71$

$\Rightarrow 71 = 7 + ( n+2-1) ( 4)$

$\Rightarrow 71 = 7 + 4n + 4$

$\Rightarrow 60 = 4n$

$\Rightarrow n = 15$

Therefore there are $15$ A.M.s

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Question 7: If $n$ $\text{A.M.s}$ are inserted between two numbers , prove that the sum of the mean equidistant from the beginning and the end is constant.

Let $A_1, A_2, \dots , A_n$ be the arithmetic means between $a$ and $b$

$\therefore a = a$  and $a_{n+2} = b$

$\therefore b = a + ( n+2-1) d$

$\Rightarrow d =$ $\frac{b-a}{n+1}$

$A_1 + A_2+ \ldots + A_n =$ $\frac{n}{2}$ $[ A_1+A_n]$

$=$ $\frac{n}{2}$ $[ A_1-d + A_n + d]$

$=$ $\frac{n}{2}$ $[a+b]$

$=$ A.M. between $a$ and $b$ which is constant.

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Question 8: If $x, y, z$ are in A.P. and $A_1$ is the $\text{A.M.}$  of $x$ and $y$  and $A_2$ is the $\text{A.M.}$ of $y$ and $z$, then prove that the $\text{A.M.s}$ of $A_1$ and $A_2$ is $y$.

Given $x, y, z$ are in A.P. Therefore $y =$ $\frac{x+z}{2}$

Now, $A_1 =\frac{x+y}{2} = \frac{x+ \big(\frac{x+z}{2} \big)}{2} = \frac{3x+z}{4}$

And $A_2 = \frac{y+z}{2} = \frac{\big(\frac{x+z}{2} \big)+z}{2} = \frac{3z+x}{4}$

Let $A_3$ be the arithmetic mean of $A_1$ and $A_2$

$\therefore A_3 = \frac{A_1+A_2}{2} = \frac{\frac{3x+z}{4} + \frac{3z+x}{4} }{2} = \frac{4x+4z}{8} = \frac{x+y}{2} = y$. Hence proved.

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Question 9: Insert five numbers between $8$ and $26$ such that the resulting sequence is an A.P..

Let $A_1, A_2, A_3, A_4, A_5$  be the 5 A.M.s between $8$ and $26$.

Therefore $8, A_1, A_2, A_3, A_4, A_5, A_6, 26$ are in A.P.

$a = 8 \ \ \ \ \ a_7 = 26$

We know $a_n = a + ( n-1) d$

$\therefore 26 = 8 + ( 7-1)(d) \Rightarrow 6d = 18 \Rightarrow d = 3$

Therefore

$A_1 = a + d = 8 +3 = 11$

$A_2 = A_1 + d = 11 +3 = 14$

$A_3 = A_2 + d = 14 +3 = 17$

$A_4 = A_3 + d = 17 +3 = 20$

$A_5 = A_4 + d = 20 +3 = 23$

Therefore the 5 A.M.s between $8$ and $26$ are $11, 14, 17, 20, 23$