Differentiation (Part-38) | S N De

 $\,18.\,$ If $\,p, q, r\,$ be in arithmetic progression, then show that the $\,p\,$th, $\,q\,$th and $\,r\,$th terms of any geometric series will be in geometric progression.

Sol. $\,\because\, p,q,r\,$ are in A.P., 

$\,\therefore 2q=p+r\rightarrow(1)$

Let $\,l\,$ be the first term of the G.P. and $\,m\,$ being the common ratio.

$\,\therefore t_p=lm^{p-1},\\~~~~~ t_q=lm^{q-1},\\~~~~~ t_r=lm^{r-1}.$

Now, $\,t_pt_r=lm^{p-1}\cdot lm^{r-1}\\~~~~~~~=l^2m^{p+r-2}\\~~~~~~~=l^2m^{2q-2}\,\,[\text{By (1)}]\\~~~~~~~=(lm^{q-1})^2\\~~~~~~~=t_q^2\rightarrow(2)$

Hence, from $\,(2),\,$ we can conclude that the $\,p\,$th, $\,q\,$th and $\,r\,$th terms of any geometric series will be in geometric progression.

$\,19.\,$ The $\,n\,$ th term of a series is $\,3^n+2n;\,$ find the sum of first $\,n\,$ terms of the series.

Sol. The sum of first $\,n\,$ terms of the series is given by :

$=(3+2)+(3^2+2 \cdot 2)+(3^3+2 \cdot 3)\\~~~~+\cdots \text{to n terms}\\=(3+3^2+3^3+\cdots +3^n)\\~~~~+2(1+2+3+\cdots +n)\\=3 \cdot \frac{3^n-1}{3-1}+2 \times \frac{n(n+1)}{2}\\=\frac 32(3^n-1)+n(n+1)\,\,\text{(ans.)}$

$\,20.\,$ If one arithmetic mean $\,A\,$ and two geometric means $\,p,q\,$ be inserted between two given numbers, then prove that $\frac{p^2}{q}+\frac{q^2}{p}=2A.$

Sol. Let the two given numbers be $\,x,y.$

$\,\therefore x+y=2A\rightarrow(1)$

Again, since $\,x,p,q,y\,$ are in G.P., 

$\,\therefore \frac px=\frac qp \Rightarrow x=\frac{p^2}{q},$

and $\,\frac qp=\frac yq \Rightarrow y=\frac{q^2}{p}.$

$\,\therefore \frac{p^2}{q}+\frac{q^2}{p}=x+y \\ \Rightarrow \frac{p^2}{q}+\frac{q^2}{p}=2A\,\,[\text{By (1)}]$

$\,21.\,$ If $\,A\,$ is the A.M.  and  $\,G\,$ is the G.M. of two unequal positive real numbers $\,p\,$ and $\,q\,$ , prove that, $\,A >G\,$ and $\,|G|>\frac{G^2}{A}.$

Sol. According to the problem, $\,A=\frac{p+q}{2}\rightarrow(1)$

and $\,G=\sqrt{pq}\rightarrow(2).$

Now, $\,\,(p-q)^2>0 \\ \Rightarrow (p+q)^2-4pq>0 \\ \Rightarrow \frac{(p+q)^2}{4}>pq \\ \Rightarrow \frac{p+q}{2}>\sqrt{pq} \\ \Rightarrow A >G.$

Again, $\,A>G \Rightarrow AG>G^2 \Rightarrow G>\frac{G^2}{A}$ 

and so $\,|G|>\frac{G^2}{A}\,[*].$

Note[*] : Since Geometric Mean of positive numbers are positive, $\,|G|=G.$ 

To download full PDF solution of Sequence and Series(Part-3), class XI for Infinite G.P. and H.P., click here . 

$\,22.\,$ Between two positive real numbers $\,a\,$ and $\,b\,$, one arithmetic mean $\,A\,$ and three geometric means $\,G_1,G_2,G_3\,\,$ are inserted. Find the value of  $\,\,\frac{(a^4+b^4)+4(G_1^4+G_3^4)+6G_2^4}{A^4}.$

Sol. Since $\,A\,$ is the arithmetic mean of $\,a\,$ and $\,b\,$, 

$\therefore \,a+b=2A \rightarrow(1)$ 

Again, since three geometric means $\,G_1,G_2,G_3\,\,$ are inserted between $\,a\,$ and $\,b\,$, so the numbers $\,a,G_1,G_2,G_3,b\,$ form G.P.

Let $\,r\,$ be the common ratio of $\,a\,$ and $\,b.$ 

So, $\,b=ar^{5-1}\,\,[\because \,b \,\text{is the 5th term}] \\ \Rightarrow b=ar^4 \\ \Rightarrow r=\left(\frac ba\right)^{\frac 14}\\ \Rightarrow r=b^{\frac 14} a^{-\frac 14},\\ G_1=ar=a^{\frac 34}b^{\frac 14}\rightarrow(2),\\ G_2=G_1r=a^{\frac 34}b^{\frac 14}b^{\frac 14}a^{-\frac 14}=a^{\frac 12}b^{\frac 12}\rightarrow(3),\\ G_3=G_2r=a^{\frac 12}b^{\frac 12}b^{\frac 14}a^{-\frac 14}=a^{\frac 14}b^{\frac 34}\rightarrow(4)\\ \therefore \frac{(a^4+b^4)+4(G_1^4+G_3^4)+6G_2^4}{A^4}\\=\frac{a^4+4G_1^4+6G_2^4+4G_3^4+b^4}{A^4}\\=\frac{a^4+4a^3b+6a^2b^2+4ab^3+b^4}{A^4}\\=\frac{(a+b)^4}{A^4}\\=\frac{(2A)^4}{A^4}\,\,[\text{By (1)}]\\=\frac{2^4 \cdot A^4}{A^4}\\=16\,\,\text{(ans.)}$

$\,23.\,$ If $\,a,b,c\,$ are in G.P. and the equation $\,ax^2+2bx+c=0\,$ and $\,dx^2+2ex+f=0\,$ have a common root, then show that $\,\,\frac da,\frac eb,\frac fc\,$ are in A.P.

Sol. If $\,a,b,c\,$ are in G.P., $\,b^2=ac\rightarrow(1)$

Now, $\,ax^2+2bx+c=0\\ \Rightarrow x=\frac{-2b \pm\sqrt{4b^2-4ac}}{2a}\\ \Rightarrow x=\frac{-2b \pm2\sqrt{b^2-ac}}{2a}\\ \Rightarrow x=\frac{-b \pm2 \times 0}{a}\,\,[\text{By (1)}]\\ \Rightarrow x=-\frac ba.$

Now, by the given problem, $\,\,-\frac ba\,$ is a root of $\,dx^2+2ex+f=0.$

So, $\,\frac{db^2}{a^2}-\frac{2be}{a}+f=0 \\ \Rightarrow \frac{dac}{a^2}-\frac{2be}{a}+f=0\,\,[\text{By (1)}] \\ \Rightarrow \frac{dc}{a}-\frac{2be}{a}+f=0\\ \Rightarrow \frac{d}{a}-\frac{2be}{ac}+\frac fc=0\\ \Rightarrow \frac{d}{a}-\frac{2be}{b^2}+\frac fc=0\,\,[\text{By (1)}]\\ \Rightarrow \frac da-\frac{2e}{b}+\frac fc=0 \\ \Rightarrow \frac da+\frac fc=2 \cdot \frac eb\rightarrow(2)$

Hence, from $\,(2),\,$ we can conclude that $\,\,\frac da,\frac eb,\frac fc\,$ are in A.P.