$\,3(i).\,$ What is the value of the H.M. between $\,(a+b)\,$ and $\,(a-b)?$
Sol. Let $\,m\,$ be the the H.M. between $\,(a+b)\,$ and $\,(a-b).$
So, $\,(a+b),m,(a-b)\,$ are in H.P. so that $\,\frac{1}{a+b},\frac 1m,\frac{1}{a-b}\,$ are in A.P.
Hence, $\,~~~~\frac 2m=\frac{1}{a+b}+\frac{1}{a-b}\\ \Rightarrow \frac 2m=\frac{a-b+a+b}{(a+b)(a-b)}\\ \Rightarrow \frac 2m=\frac{2a}{a^2-b^2}\\ \Rightarrow \frac 1m=\frac{a}{a^2-b^2}\\ \Rightarrow m=\frac{a^2-b^2}{a}\,\,\text{(ans.)}$
$\,3(ii)\,$ Insert two H.M.s between $\,\frac 13\,$ and $\,\frac{1}{24}.$
Sol. Let $\,m,\,n\,$ be two H.M.s between $\,\frac 13\,$ and $\,\frac{1}{24}.$
So, $\,3,\frac 1m,\frac 1n, 24\,$ form an A.P. where the first term $(a)=3,\,$ fourth term $(t_4)=24.$
Now, $\,~~~~~~t_4=a+(4-1)d\\~~~~~~~~[\text{where d is common difference}]\\ \Rightarrow 24=3+3d\\ \Rightarrow 24=3(1+d)\\ \Rightarrow 1+d=\frac{24}{3}\\ \Rightarrow d=8-1=7.$
Now, $\,t_2=a+(2-1)d \\ \Rightarrow \frac 1m=3+d=3+7=10 \\ \Rightarrow m=\frac{1}{10}.$
Again, $\,t_3=a+(3-1)d \\ \Rightarrow \frac 1n=3+2d=3+2 \times 7=17 \\ \Rightarrow n=\frac{1}{17}.$
Hence, two H.M.s between $\,\frac 13\,$ and $\,\frac{1}{24}\,\,$ are $\,\frac{1}{10}\,$ and $\,\frac{1}{17}.$
$\,3(iii)\,$ Insert eight H.M.s between $\,(-6)\,$ and $\,\frac{3}{13}.$
Sol. Let $\,m_1,m_2,\cdots,m_8\,$ be the eight H.M.s between $\,(-6)\,$ and $\,\frac{3}{13}.$
Hence, $\,-\frac 16,\frac{1}{m_1},\frac{1}{m_2},\cdots,\frac{1}{m_8}, \frac{13}{3} \,$ form an A.P. Let $\,d\,$ be the common difference of the A.P.
Now, first term $\,(a)=-\frac 16,\\~~t_{10}=\frac{13}{3}\\ \Rightarrow -\frac 16+(10-1)d=\frac{13}{3}\\ \Rightarrow 9d=\frac{13}{3}+\frac 16 \\ \Rightarrow d=\frac 19 \times \left(\frac{26+1}{6}\right)\\ \therefore d=\frac 19 \times \frac{27}{6}=\frac 12.$
So, $\,t_2=a+d=-\frac 16+\frac 12=\frac{-1+3}{6}\\ \Rightarrow \frac{1}{m_1}=\frac 26=\frac 13 \\ \therefore m_1=3.$
$\,t_3=a+2d=-\frac 16+2 \cdot \frac 12=-\frac 16+1\\ \Rightarrow \frac{1}{m_2}=\frac 56 \\ \therefore m_2=\frac 65. $
$\,t_4=a+3d=-\frac 16+3 \cdot \frac 12=-\frac 16+\frac 32\\ \Rightarrow \frac{1}{m_3}=\frac 43 \\ \therefore m_3=\frac 34. $
$\,t_5=a+4d=-\frac 16+4 \cdot \frac 12=-\frac 16+2\\ \Rightarrow \frac{1}{m_4}=\frac{11}{6} \\ \therefore m_4=\frac{6}{11}. $
$\,t_6=a+5d=-\frac 16+5 \cdot \frac 12=-\frac 16+\frac 52\\ \Rightarrow \frac{1}{m_5}=\frac{7}{3} \\ \therefore m_5=\frac{3}{7}. $
$\,t_7=a+6d=-\frac 16+6 \cdot \frac 12=-\frac 16+3\\ \Rightarrow \frac{1}{m_6}=\frac{17}{6} \\ \therefore m_6=\frac{6}{17}. $
$\,t_8=a+7d=-\frac 16+7 \cdot \frac 12=-\frac 16+\frac 72\\ \Rightarrow \frac{1}{m_7}=\frac{10}{3} \\ \therefore m_7=\frac{3}{10}. $
$\,t_9=a+8d=-\frac 16+8 \cdot \frac 12=-\frac 16+4\\ \Rightarrow \frac{1}{m_8}=\frac{23}{6} \\ \therefore m_8=\frac{6}{23}. $
Hence, eight H.M.s between $\,(-6)\,$ and $\,\frac{3}{13}\,$ are
$\,3,\frac 65, \frac 34, \frac{6}{11},\frac 37,\frac{6}{17},\frac{3}{10},\frac{6}{23}.$
$\,4(i).\,$ If the G.M. and H.M. of two numbers be $\,15\,$ and $\,9\,$ respectively, find the numbers.
Sol. Let $\,a\,$ and $\,b\,$ be two numbers.
Since $\,15\,$ is the G.M. between $\,a\,$ and $\,b\,$,
so $\,\sqrt{ab}=15 \Rightarrow ab=15^2=225\rightarrow(1)$
Since $\,9\,$ is the H.M. between $\,a\,$ and $\,b\,$,
$\,\frac 1a,\frac 19,\frac 1b\,$ are in A.P.
So, $\,\frac 1a+\frac 1b=\frac 29 \\ \Rightarrow \frac{a+b}{ab}=\frac 29 \\ \Rightarrow 9(a+b)=2ab =2\times 225 \\ \Rightarrow 9\left(a+\frac{225}{a}\right)=450 ~~~[\text{By (1)}]\\ \Rightarrow \frac{a^2+225}{a}=\frac{450}{9}=50 \\ \Rightarrow a^2-50a+225=0 \\ \Rightarrow a^2-5a-45a+225=0 \\ \Rightarrow a(a-5)-45(a-5)=0 \\ \Rightarrow (a-5)(a-45)=0 \\ \Rightarrow a=5,45.$
If $\,a=5,\,\,b=\frac{225}{a}=\frac{225}{5}=45, \\ a=45,\,\,b=\frac{225}{a}=\frac{225}{45}=5.$
Hence, the numbers are $\,5\,$ and $\,45.$
$\,4(ii)\,$ If the A.M. and H.M. of two numbers are $\,15\,$ and $\,9\frac 35\,$ respectively, determine the numbers.
Sol. Let $\,a\,$ and $\,b\,$ be two numbers such that the A.M. and H.M. of these two numbers are $\,15\,$ and $\,9\frac 35\,$ respectively.
Now, $\,~~~\frac{a+b}{2}=15 \\ \Rightarrow a+b=30 \rightarrow(1)$
Again, by the given condition, $\,\frac 1a,\frac{1}{9\frac 35}=\frac{5}{48},\,\frac 1b\,$ are in A.P.
So, $\,~~~~\frac 1a+\frac 1b=2\times \frac{5}{48}\\ \Rightarrow \frac{b+a}{ab}=\frac{5}{24} \\ \Rightarrow \frac{30}{ab}=\frac{5}{24} ~~[\text{By (1)}]\\ \Rightarrow ab=144 \rightarrow(2)$
Hence, from $\,(1),\,(2)\,$ we get,
$\,a+\frac{144}{a}=30 \\ \Rightarrow a^2-30a+144=0 \\ \Rightarrow a^2-6a-24a+144=0 \\ \Rightarrow a(a-6)-24(a-6)=0 \\ \Rightarrow (a-6)(a-24)=0 \\ \Rightarrow a=6,24.$
When $\,a=24,\,b=30-24=6,\\ ~a=6,\,b=30-6=24.$
Hence, two numbers are $\,6,24.$
To download full PDF solution of Sequence and Series(Part-3), class XI for Infinite G.P. and H.P., click here .
$\,4(iii)\,$ If the ratio of G.M. of two numbers to their H.M. is $\,5 :4,\,$ prove that the numbers are in the ratio of $\,4:1.$
Sol. Let $\,a,\,b\,$ be two numbers so that G.M. of $\,a,b\,$ is $=\sqrt{ab}\,$ and H.M. of these two numbers is $=\frac{2ab}{a+b}.$
So, $~~~~\frac{\sqrt{ab}}{\frac{2ab}{a+b}}=\frac 54 \\ \Rightarrow \frac{(a+b)\sqrt{ab}}{2ab}=\frac 54 \\ \Rightarrow \frac{a+b}{\sqrt{ab}}=\frac 52 \\ \Rightarrow \left(\frac{a+b}{\sqrt{ab}}\right)^2=\left(\frac 52\right)^2 \\ \Rightarrow \frac{a^2+2ab+b^2}{ab}=\frac{25}{4} \\ \Rightarrow 4a^2+8ab+4b^2=25ab \\ \Rightarrow 4a^2-17ab+4b^2=0 \\ \Rightarrow 4a^2-16ab-ab+4b^2=0 \\ \Rightarrow 4a(a-4b)-b(a-4b)=0 \\ \Rightarrow (a-4b)(4a-b)=0 \\ \Rightarrow a=4b\rightarrow(1) \\ \text{or,}~~4a=b\rightarrow(2)$
Hence, from $\,(1),\,(2)\,$ we can conclude that the numbers are in the ratio of $\,4:1.$
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