\displaystyle \textbf{Question 1: }\text{Find the A.M. between:}
\displaystyle \text{i) }7\text{ and }13\qquad\text{ii) }12\text{ and }-8\qquad\text{iii) }(x-y)\text{ and }(x+y).
\displaystyle \text{Answer:}
\displaystyle \text{i) Let }A\text{ be the arithmetic mean of }7\text{ and }13.
\displaystyle \therefore 7,A,13\text{ are in A.P.}
\displaystyle A-7=13-A
\displaystyle 2A=20
\displaystyle \therefore A=\frac{7+13}{2}=10
\displaystyle \therefore \text{The arithmetic mean is }10.
\displaystyle \text{ii) Let }A\text{ be the arithmetic mean of }12\text{ and }-8.
\displaystyle \therefore 12,A,-8\text{ are in A.P.}
\displaystyle A-12=-8-A
\displaystyle 2A=4
\displaystyle \therefore A=\frac{12+(-8)}{2}=2
\displaystyle \therefore \text{The arithmetic mean is }2.
\displaystyle \text{iii) Let }A\text{ be the arithmetic mean of }(x-y)\text{ and }(x+y).
\displaystyle \therefore (x-y),A,(x+y)\text{ are in A.P.}
\displaystyle A-(x-y)=(x+y)-A
\displaystyle 2A=(x-y)+(x+y)
\displaystyle \therefore A=\frac{(x-y)+(x+y)}{2}=x
\displaystyle \therefore \text{The arithmetic mean is }x.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Insert }4\text{ A.M.s between }4\text{ and }19.
\displaystyle \text{Answer:}
\displaystyle \text{Let }A_1,A_2,A_3,A_4\text{ be the four arithmetic means.}
\displaystyle \therefore 4,A_1,A_2,A_3,A_4,19\text{ are in A.P.}
\displaystyle a=4,\qquad a_6=19
\displaystyle a_n=a+(n-1)d
\displaystyle 19=4+(6-1)d
\displaystyle 5d=15
\displaystyle \therefore d=3
\displaystyle A_1=a+d=4+3=7
\displaystyle A_2=a+2d=4+2(3)=10
\displaystyle A_3=a+3d=4+3(3)=13
\displaystyle A_4=a+4d=4+4(3)=16
\displaystyle \therefore \text{The four arithmetic means are }7,10,13,16.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Insert }7\text{ A.M.s between }2\text{ and }17.
\displaystyle \text{Answer:}
\displaystyle \text{Let }A_1,A_2,A_3,A_4,A_5,A_6,A_7\text{ be the seven arithmetic means.}
\displaystyle \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,\qquad a_9=17
\displaystyle a_n=a+(n-1)d
\displaystyle 17=2+(9-1)d
\displaystyle 15=8d
\displaystyle \therefore d=\frac{15}{8}
\displaystyle A_1=a+d=2+\frac{15}{8}=\frac{31}{8}
\displaystyle A_2=a+2d=2+\frac{30}{8}=\frac{46}{8}
\displaystyle A_3=a+3d=2+\frac{45}{8}=\frac{61}{8}
\displaystyle A_4=a+4d=2+\frac{60}{8}=\frac{76}{8}
\displaystyle A_5=a+5d=2+\frac{75}{8}=\frac{91}{8}
\displaystyle A_6=a+6d=2+\frac{90}{8}=\frac{106}{8}
\displaystyle A_7=a+7d=2+\frac{105}{8}=\frac{121}{8}
\displaystyle \therefore \text{The seven arithmetic means between }2\text{ and }17\text{ are }
\displaystyle \frac{31}{8},\frac{46}{8},\frac{61}{8},\frac{76}{8},\frac{91}{8},\frac{106}{8},\frac{121}{8}.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Insert }6\text{ A.M.s between }15\text{ and }-13.
\displaystyle \text{Answer:}
\displaystyle \text{Let }A_1,A_2,A_3,A_4,A_5,A_6\text{ be the six arithmetic means.}
\displaystyle \therefore 15,A_1,A_2,A_3,A_4,A_5,A_6,-13\text{ are in A.P.}
\displaystyle a=15,\qquad a_8=-13
\displaystyle a_n=a+(n-1)d
\displaystyle -13=15+(8-1)d
\displaystyle 7d=-28
\displaystyle \therefore d=-4
\displaystyle A_1=a+d=15-4=11
\displaystyle A_2=a+2d=15-8=7
\displaystyle A_3=a+3d=15-12=3
\displaystyle A_4=a+4d=15-16=-1
\displaystyle A_5=a+5d=15-20=-5
\displaystyle A_6=a+6d=15-24=-9
\displaystyle \therefore \text{The six arithmetic means between }15\text{ and }-13\text{ are }11,7,3,-1,-5,-9.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{There are }n\text{ A.M.s between }3\text{ and }17.\text{ The ratio of the last mean}
\displaystyle \text{to the first mean is }3:1.\text{ Find the value of }n.
\displaystyle \text{Answer:}
\displaystyle \text{Let }A_1,A_2,\ldots,A_n\text{ be the }n\text{ arithmetic means between }3\text{ and }17.
\displaystyle \therefore 3,A_1,A_2,\ldots,A_n,17\text{ are in A.P.}
\displaystyle a=3,\qquad a_{n+2}=17
\displaystyle a_k=a+(k-1)d
\displaystyle 17=3+\{(n+2)-1\}d
\displaystyle 14=(n+1)d
\displaystyle \therefore d=\frac{14}{n+1}
\displaystyle A_1=a+d
\displaystyle =3+\frac{14}{n+1}
\displaystyle =\frac{3n+17}{n+1}
\displaystyle A_n=a+nd
\displaystyle =3+n\left(\frac{14}{n+1}\right)
\displaystyle =\frac{17n+3}{n+1}
\displaystyle \text{Given, }\frac{A_n}{A_1}=\frac{3}{1}
\displaystyle \frac{\frac{17n+3}{n+1}}{\frac{3n+17}{n+1}}=\frac{3}{1}
\displaystyle \frac{17n+3}{3n+17}=3
\displaystyle 17n+3=9n+51
\displaystyle 8n=48
\displaystyle \therefore n=6
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{Insert A.M.s between }7\text{ and }71\text{ such that the }5^{\text{th}}\text{ A.M. is }27.
\displaystyle \text{Find the number of A.M.s.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }A_1,A_2,\ldots,A_n\text{ be the arithmetic means between }7\text{ and }71.
\displaystyle \therefore 7,A_1,A_2,\ldots,A_n,71\text{ are in A.P.}
\displaystyle \text{The }5^{\text{th}}\text{ A.M. is the }6^{\text{th}}\text{ term of the A.P.}
\displaystyle a=7,\qquad a_6=27
\displaystyle a_k=a+(k-1)d
\displaystyle 27=7+(6-1)d
\displaystyle 20=5d
\displaystyle \therefore d=4
\displaystyle \text{Also, }a_{n+2}=71
\displaystyle 71=7+\{(n+2)-1\}(4)
\displaystyle 71=7+4(n+1)
\displaystyle 64=4(n+1)
\displaystyle n+1=16
\displaystyle \therefore n=15
\displaystyle \therefore \text{The number of arithmetic means is }15.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{If }n\text{ A.M.s are inserted between two numbers, prove that the sum of}
\displaystyle \text{the means equidistant from the beginning and the end is constant.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }A_1,A_2,\ldots,A_n\text{ be the }n\text{ arithmetic means inserted between }a\text{ and }b.
\displaystyle \therefore a,A_1,A_2,\ldots,A_n,b\text{ are in A.P.}
\displaystyle \text{Let the common difference be }d.
\displaystyle A_r=a+rd
\displaystyle \text{The mean equidistant from the end is }A_{n-r+1}.
\displaystyle A_{n-r+1}=a+(n-r+1)d
\displaystyle \therefore A_r+A_{n-r+1}
\displaystyle =a+rd+a+(n-r+1)d
\displaystyle =2a+(n+1)d
\displaystyle \text{Since }b=a+(n+1)d,
\displaystyle (n+1)d=b-a
\displaystyle \therefore A_r+A_{n-r+1}=2a+(b-a)
\displaystyle =a+b
\displaystyle \therefore \text{The sum of any two means equidistant from the beginning and the end is }a+b,
\displaystyle \text{which is constant.}
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{If }x,y,z\text{ are in A.P., }A_1\text{ is the A.M. of }x\text{ and }y,\text{ and }A_2
\displaystyle \text{is the A.M. of }y\text{ and }z,\text{ prove that the A.M. of }A_1\text{ and }A_2\text{ is }y.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }x,y,z\text{ are in A.P.}
\displaystyle \therefore 2y=x+z
\displaystyle \therefore y=\frac{x+z}{2}
\displaystyle A_1=\frac{x+y}{2}
\displaystyle =\frac{x+\frac{x+z}{2}}{2}
\displaystyle =\frac{3x+z}{4}
\displaystyle A_2=\frac{y+z}{2}
\displaystyle =\frac{\frac{x+z}{2}+z}{2}
\displaystyle =\frac{x+3z}{4}
\displaystyle \text{Let }A_3\text{ be the arithmetic mean of }A_1\text{ and }A_2.
\displaystyle A_3=\frac{A_1+A_2}{2}
\displaystyle =\frac{\frac{3x+z}{4}+\frac{x+3z}{4}}{2}
\displaystyle =\frac{4x+4z}{8}
\displaystyle =\frac{x+z}{2}
\displaystyle \therefore A_3=y
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{Insert five numbers between }8\text{ and }26\text{ such that the resulting sequence}
\displaystyle \text{is an A.P.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }A_1,A_2,A_3,A_4,A_5\text{ be the five arithmetic means between }8\text{ and }26.
\displaystyle \therefore 8,A_1,A_2,A_3,A_4,A_5,26\text{ are in A.P.}
\displaystyle a=8,\qquad a_7=26
\displaystyle a_n=a+(n-1)d
\displaystyle 26=8+(7-1)d
\displaystyle 18=6d
\displaystyle \therefore d=3
\displaystyle A_1=a+d=8+3=11
\displaystyle A_2=a+2d=8+2(3)=14
\displaystyle A_3=a+3d=8+3(3)=17
\displaystyle A_4=a+4d=8+4(3)=20
\displaystyle A_5=a+5d=8+5(3)=23
\displaystyle \therefore \text{The five numbers are }11,14,17,20,23.
\displaystyle \\

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