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

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

Answer:

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 .

Answer:

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 .

Answer:

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

Answer:

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 .

Answer:

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} .

Answer:

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.

Answer:

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 .

Answer:

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..

Answer:

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