Differentiation (Part-38) | S N De

 $\,24.\,$ If $\,a\,$ is A.M. of two positive numbers $\,b\,$ and $\,c\,$ and two G.M.'s between $\,b\,$ and $\,c\,$ are $\,G_1\,$ and $\,G_2\,$, prove that $\,G_1^3+G_2^3=2abc.$

Sol. If $\,a\,$ is A.M. of two positive numbers $\,b\,$ and $\,c\,$,

$\,b+c=2a\rightarrow(1)$

Again, since two G.M.'s between $\,b\,$ and $\,c\,$ are $\,G_1\,$ and $\,G_2\,$, 

$\,b,G_1,G_2,c\,$ form G.P. with common ratio $\,r\,$, say.

$\,\therefore c=t_4=br^{4-1},\,[\text{where }\,\,t_n\,\, \text{denotes n-th term.}]\\ \Rightarrow c=br^3 \\ \Rightarrow r=(c/b)^{\frac 13}=c^{\frac 13}b^{-\frac 13}. \\ \therefore G_1=br=b \cdot c^{\frac 13}b^{-\frac 13}=b^{\frac 23}c^{\frac 13}\\ \text{and}\,\,G_2=G_1r=b^{\frac 23}c^{\frac 13}c^{\frac 13}b^{-\frac 13}=c^{\frac 23}b^{\frac 13}$

Now, $\,\,G_1^3+G_2^3=b^2c+bc^2\\ \Rightarrow G_1^3+G_2^3=bc(b+c)\\ \Rightarrow G_1^3+G_2^3=2abc\,\,[\text{By (1)}]$

$\,25.\,$ If $\,G\,$ is the G.M. between two positive numbers $\,a\,$ and $\,b$ , show that $\,\frac{1}{G^2-a^2}+\frac{1}{G^2-b^2}=\frac{1}{G^2}.$

Sol.  $\,\because \, G\,$ is  the G.M. between two positive numbers $\,a\,$ and $\,b$ , 

$\,\therefore G^2=ab \rightarrow(1)$

Now, $\,\frac{1}{G^2-a^2}+\frac{1}{G^2-b^2}\\=\frac{1}{ab-a^2}+\frac{1}{ab-b^2}\\=\frac{1}{-a(a-b)}+\frac{1}{b(a-b)}\\=\frac{1}{(a-b)}\left[\frac 1b-\frac 1a\right]\\=\frac{1}{(a-b)}\times \frac{a-b}{ba}\\=\frac{1}{ab}\\=\frac{1}{G^2}\,\,\,[\text{By (1)}]\,\,\text{(showed)}$

$\,27.\,$ If sum of the square of three different terms in G.P. is $\,S^2\,$ and their sum is $\,\alpha S,\,$ then show that $\,\,\frac 13<\alpha^2<3.$

Sol. Let  three different terms in G.P. are $\,\frac ar,a,ar.$

Now, since sum of the square of three different terms in G.P. is $\,S^2\,$ ,

$\,(\frac ar)^2+a^2+a^2r^2=S^2\\ \Rightarrow a^2(\frac{1}{r^2}+1+r^2)=S^2\rightarrow(1)$

Again, according to the problem,

$\,\frac ar+a+ar=\alpha S\\ \Rightarrow a(\frac 1r+1+r)=\alpha S\\ \Rightarrow a(y+1)=\alpha S\,\,[\text{let}\,\,r+\frac 1r=y]\\ \Rightarrow a^2(y+1)^2=\alpha^2S^2\rightarrow(2)$

Again, from $\,(1),\,$ we have,

$\,a^2[(r+\frac 1r)^2-2 \cdot r \cdot \frac 1r+1]=S^2\\ \Rightarrow a^2(y^2-1)=S^2\rightarrow(3)$

So, from $\,(2),\,(3)\,$ we get,

$\,a^2(y+1)^2=\alpha^2 \cdot a^2(y^2-1) \\ \Rightarrow \alpha^2=\frac{(y+1)^2}{(y+1)(y-1)}\rightarrow(4)\\ \Rightarrow \alpha^2=\frac{y+1}{y-1}=\frac{y-1+2}{y-1}\\ \Rightarrow \alpha^2=1+\frac{2}{y-1}$

From $\,(4),\,$ we can say that $\, y \neq \pm1\,$ and so, 

$\,y \in (-\infty,-2) \cup (2,\infty) \\ \Rightarrow y-1 \in  (-\infty,-3) \cup (1,\infty)  \\ \Rightarrow \frac{1}{y-1} \in (-\frac 13,0) \cup (0,1)   \\ \Rightarrow \frac{2}{y-1} \in (-\frac 23,0) \cup (0,2)  \\ \Rightarrow 1+\frac{1}{y-1} \in (\frac 13,1) \cup (1,3) \\ \therefore \alpha^2 \in (\frac 13,1) \cup (1,3)\rightarrow(5)  $

Hence, from $\,(5),\,$ we can conclude that $\,\,\frac 13<\alpha^2<3.$

To download full PDF solution of Arithmetic Progression (SEQUENCE AND SERIES) of S.N.De Math solution, click here.

$\,28.\,$ Show that sum of $\,n\,$ G.M.s between $\,a\,$ and $\,b\,$ is :

$\,\left(\frac{ab^{\frac{1}{n+1}}-ba^{\frac{1}{n+1}}}{a^{\frac{1}{n+1}}-b^{\frac{1}{n+1}}}\right)$

Sol. Let the $\,n\,$ G.M.s between $\,a\,$ and $\,b\,$ be $\,x_1,x_2,\cdots,x_n.$

So, $\,\, a,x_1,x_2,\cdots,x_n,b\,$ are in G.P. with common ratio $\,r,\,\text{(say).}$

$\,\therefore b=ar^{n+1}=t_{n+1}\\ \Rightarrow r=(b/a)^{\frac{1}{n+1}}\rightarrow(1)\\ \therefore x_1=ar=a\cdot (b/a)^{\frac{1}{n+1}}$

Now, $\,x_1+x_2+\cdots +x_n\\=x_1 \cdot \frac{r^n-1}{r-1}\\=a\cdot (b/a)^{\frac{1}{n+1}}\cdot \frac{(b/a)^{\frac{n}{n+1}}-1}{(b/a)^{\frac{1}{n+1}}-1}\\=\frac{a \cdot \frac ba -a \cdot (b/a)^{\frac{1}{n+1}}}{(b/a)^{\frac{1}{n+1}}-1}\\=\frac{b-a  \cdot (b/a)^{\frac{1}{n+1}}}{(b/a)^{\frac{1}{n+1}}-1}\\=\frac{b \cdot a^{\frac{1}{n+1}}-a \cdot b^{\frac{1}{n+1}}}{b^{\frac{1}{n+1}}-a^{\frac{1}{n+1}}}\\=\frac{ab^{\frac{1}{n+1}}-ba^{\frac{1}{n+1}}}{a^{\frac{1}{n+1}}-b^{\frac{1}{n+1}}}\,\,\,\text{(showed)}$

$\,29.\,$ Evaluate :

$\,\sum_{k=1}^7 \left[(\frac 15)^{2k+1} . (7)^{k-1}\right]$

Sol. $\,\sum_{k=1}^7 \left[(\frac 15)^{2k+1} . (7)^{k-1}\right]\\=\frac{1}{7 \times 5}\sum_{k=1}^7 (\frac 15)^{2k}.(\sqrt 7)^{2k}\\=\frac{1}{7 \times 5}.\sum_{k=1}^7\left(\frac{\sqrt7}{5}\right)^{2k}\\=\frac{1}{7 \times 5} \left[(\frac{\sqrt7}{5})^2+(\frac{\sqrt7}{5})^4+\cdots+(\frac{\sqrt7}{5})^{14}\right]\\=\frac{1}{7 \times 5} \left(\frac{\sqrt7}{5}\right)^2.\frac{\left[\left(\frac{\sqrt7}{5}\right)^2\right]^7-1}{\left(\frac{\sqrt7}{5}\right)^2-1}\\=\frac{1}{7 \times 5} \times \frac{7}{25}.\frac{\frac{7^7}{5^{14}}-1}{\frac{7}{25}-1}\\=\frac{1}{7 \times 5} \times \frac{\frac{7^7-5^{14}}{5^{14}}}{\frac{-18}{5^2}}\\=\frac{1}{7 \times 5}\times \frac{5^{14}-7^7}{5^{14}}\,\,\,\text{(ans.)}$