Class 11: Arithmetic Progressions – Exercise 19.6 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

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

$\displaystyle \text{i) } 7 \text{ and } 13 \hspace{1,0cm} \text{ii) } 12 \text{ and } -8 \hspace{1,0cm} \text{iii) } (x-y) \text{ and } (x + y) .$

Answer:

$\displaystyle \text{i) } \text{Let } A \text{ be the Arithmetic Mean of } 7 \text{ and } 13$ .

$\displaystyle \text{Therefore } 7, A, 13 \text{ are in A.P. }$

$\displaystyle \Rightarrow A - 7 = 13 - A$

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

$\displaystyle \text{Hence the Arithmetic Mean of } 7 \text{ and } 13 \text{ is } 10$

$\displaystyle \text{ii) } \text{Let } A \text{ be the Arithmetic Mean of } 12 \text{ and } -8$

$\displaystyle \text{Therefore } 12, A, -8 \text{ are in A.P. }$

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

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

$\displaystyle \text{Hence the Arithmetic Mean of } 12 \text{ and } -8 \text{ is } 2$

$\displaystyle \text{iii) } \text{Let } A \text{ be the Arithmetic Mean of } (x-y) \text{ and } (x+y)$

$\displaystyle \text{Therefore } (x-y), A, (x+y) \text{ are in A.P. }$

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

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

$\displaystyle \text{Hence the Arithmetic Mean of } (x-y) \text{ and } (x+y) \text{ is } x$

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

Answer:

$\displaystyle \text{Let } A_1, A_2, A_3, A_4,$ be the 4 A.M.s between $\displaystyle 4 \text{ and } 19$

$\displaystyle \text{Therefore } 4, A_1, A_2, A_3, A_4, 19 \text{ are in A.P. }$

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

$\displaystyle \text{We know } a_n = a + ( n-1) d$

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

Therefore

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

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

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

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

Therefore the 4 A.M.s between $\displaystyle 4 \text{ and } 19$ are $\displaystyle 7, 10, 13 ,16$

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

Answer:

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

$\displaystyle \text{Therefore } 2, A_1, A_2, A_3, A_4, A_5, A_6, A_7, 17 \text{ are in A.P. }$

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

$\displaystyle \text{We know } a_n = a + ( n-1) d$

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

Therefore

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

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

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

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

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

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

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

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

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

Answer:

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

$\displaystyle \text{Therefore } 15, A_1, A_2, A_3, A_4, A_5, A_6, -13 \text{ are in A.P. }$

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

$\displaystyle \text{We know } a_n = a + ( n-1) d$

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

Therefore

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

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

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

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

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

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

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

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

Answer:

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

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

$\displaystyle \text{We know } a_n = a + ( n-1) d$

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

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

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

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

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

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

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

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

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

$\displaystyle \Rightarrow 8n = 48$

$\displaystyle \Rightarrow n = 6$

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

Answer:

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

$\displaystyle \text{Given } a = 7 \ \ a_6 = 27$

$\displaystyle \therefore a + 5d = 27$

$\displaystyle \Rightarrow 5d = 27 - 7$

$\displaystyle \Rightarrow d = 4$

$\displaystyle A_{n+2} = 71$

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

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

$\displaystyle \Rightarrow 60 = 4n$

$\displaystyle \Rightarrow n = 15$

Therefore there are $\displaystyle 15 \text{ A.M.s }$

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

Answer:

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

$\displaystyle \therefore a = a \text{ and } a_{n+2} = b$

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

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

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

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

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

$\displaystyle = \text{ A.M. between } a \text{ and } b$ which is constant.

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