Differentiation (Part-38) | S N De

 

$\,1.\,$ If $\,5,x,y,z,405\,\,$ are the first five terms of a G.P., find the values of $\,\,x,y,z.$

Sol.  Clearly in the given G.P., first term $(a)=5\,$ and the fifth term $(t_5)=405.$

Now, $\,t_5=a \cdot r^{5-1}\,\,~[~r \,~\text{being the common ratio}]\\ \Rightarrow 405=5 \cdot r^4 \\ \Rightarrow r^4=\frac{405}{5}=81 \\ \Rightarrow r=\sqrt[4]{81}=3$

Hence, $\,x=t_2=5 \cdot 3^{2-1}=5 \times 3=15,\\~y=t_3=5 \cdot 3^{3-1}=5 \times 3^2=45,\\ ~z=t_4=5 \cdot 3^{4-1}=5 \times 3^3=135.$

$\,2.\,$  The second term of a G.P. is $\,b\,$ and the common ratio is $\,r\,$. If the product of the first three terms of this G.P. is $\,64\,$, find $\,b\,$.

Sol. Since the second term of a G.P. is $\,b\,$ and the common ratio is $\,r\,$, the other two terms are $\,\,\frac br\,\text{and}\,\,br\,$ where $\,t_1=\frac br,\,$ and $\,t_3=br.$ 

Now, according to the problem,

$\,\frac br .b.(br)=64 \\ \Rightarrow b^3=64\\ \Rightarrow b=\sqrt[3]{64}=4\,\text{(ans.)}$

$\,3(i).\,$ Find the $\,10\,$ th and $\,p\,$th  terms of the G.P.

$~~~~~~~~~~~\{2, 6, 18, 54, \cdots \}$

Sol. Here, $\,a=t_1=2,\\ ~r=\frac 62=3.$

So, $~~\,t_{10}=ar^{10-1}=2 \cdot 3^{10-1}=2 \times 3^9,\\~~~~ t_p=ar^{p-1}=2 \cdot 3^{p-1}$

$\,3(ii)\,\,$  Find the $\,9\,$th and $\,q\,$ th terms of the G.P.

$\,~~~~~~~~~~~\{4, -8, 16, -32,\cdots \}$

Sol. Here, $\,a=t_1=4,\\ ~r=\frac{-8}{4}=-2.$

So, $~~\,t_{9}=ar^{9-1}=4 \cdot (-2)^{9-1}=4 \times 2^8=1024,\\~~~~ t_q=ar^{q-1}=4 \cdot (-2)^{q-1}\\~~~~~~~~~~~~~~~~~~~~~~=2^2 \cdot (-1)^{q-1} \cdot 2^{q-1}\\~~~~~~~~~~~~~~~~~~~~~~=(-1)^{q-1} \cdot 2^{q-1+2}\\~~~~~~~~~~~~~~~~~~~~~~=(-1)^{q-1} \cdot 2^{q+1}.$

$\,3(iii)\,\,$ Find the $\,14\,$ th and $\,n\,$ th terms of the G.P.

$\,~~~~~~~~~~\{\sqrt5,1,\frac{1}{\sqrt5}, \frac 15, \cdots\}$

Sol. Here, $\,a=t_1=\sqrt5,\\ ~r=\frac{1}{\sqrt5}.$

So, $~~\,t_{14}=ar^{14-1}\\~~~~~~~=\sqrt5 \cdot (1/\sqrt5)^{14-1}\\~~~~~~~=\sqrt5 \times \frac{1}{(\sqrt5)^{13}}\\~~~~~~~=\frac{1}{(\sqrt5)^{12}}\\~~~~~~~=\frac{1}{5^6}\\~~~~~~~=5^{-6}\,\,\text{(ans.)},\\~~~~ t_n=ar^{n-1}\\~~~~~~~=\sqrt5 \cdot (1/\sqrt5)^{n-1}\\~~~~~~~=5^{\frac 12} \cdot 5^{-\frac 12(n-1)}\\~~~~~~~=5^{\frac 12(1-n+1)}\\~~~~~~~=5^{\frac 12(2-n)}\\~~~~~~~=5^{1-\frac n2}\,\,\text{(ans.)}$

$\,4(i)\,$ If the third term of a G.P. is the square of the first term and its fifth term is $\,729\,$, find the G.P.

Sol. Let the G.P. be $\,a,ar,ar^2,\cdots.$

Now, by the given condition,

$\,t_1^2=t_3\\ \Rightarrow a^2=ar^{3-1}\\ \Rightarrow a^2=ar^2 \\ \therefore a=r^2\rightarrow(1)$

Again, $\,t_5=729\,~~[\text{(given)}] \\ \Rightarrow ar^{4-1}=729 \\ \Rightarrow ar^4=729 \\ \Rightarrow r^2 \cdot r^4=729 \,\,[\text{By (1)}] \\ \Rightarrow r^6=3^6 \\ \Rightarrow r=3.$

Hence, $\,a=r^2=3^2=9.$

So, for $\,a=9,\,r=3\,\,$ the G.P. is : $\,\{9, 9 \cdot 3, 9 \cdot 3^2,\cdots\}\\=\{9,27,81, \cdots\}$

Again, for $\,a=9,\,r=-3\,$ we get the G.P.

$\,\,=\{9,9 \cdot (-3), 9\cdot (-3)^2, 9 \cdot (-3)^3, \cdots\}\\=\{9,-27,81,-243,\cdots\}$

 $\,4(ii)\,$  If the $\,(p+q)\,$th term of a G.P. is $\,a\,$ and $\,(p-q)\,$ th term is $\,b\,$, determine its $\,p\,$th term.

Sol. $~~~t_{p+q}=ar^{p+q-1}\\ \Rightarrow a=ar^{p+q-1}\rightarrow(1)$

Again, $~~\,t_{p-q}=ar^{p-q-1}\\ \Rightarrow b=ar^{p-q-1}\rightarrow(2)$

Now, from $\,(1),\,(2)\,$ we get,

$\,ab=(ar^{p+q-1})(ar^{p-q-1})\\ \Rightarrow ab=a^2r^{(p+q-1)+(p-q-1)}\\~~~~~~~~~=a^2r^{2p-2}\\~~~~~~~~~=(ar^{p-1})^2 \\ \Rightarrow \pm\sqrt{ab}=ar^{p-1}=t_p.$

Hence, the $\,p\,$th term of the G.P. is : $\,\pm\sqrt{ab}.$

$\,4(iii)\,$ If the $\,n\,$th term of a G.P. be $\,p\,$, then show that the product of its first $\,(2n-1)\,$ terms is $\,p^{2n-1}$.

Sol. Let the first term of the G.P. be $\,a\,$ and the common ratio being $\,r.$

Let the $\,n\,$th term of a G.P., denoted by $\,t_n\,$ and is given by 

$\,t_n=p \Rightarrow ar^{n-1}=p\rightarrow(1)$

Now, the product of first $\,(2n-1)\,$ terms is :

$=a \cdot (ar) \cdot (ar^2) \cdot \cdots (ar^{(2n-1)-1})\\=a^{(2n-1)} \cdot r^{1+2+(2n-2)}\\=a^{(2n-1)} \cdot r^{\frac{(2n-2)(2n-2+1)}{2}}\\=a^{(2n-1)}\cdot r^{(n-1)(2n-1)}\\=[ar^{n-1}]^{(2n-1)}\\=p^{2n-1}\,\,\,\text{[By (1)]}~~~\text{(showed)}$

$\,5(i)\,$  Which term of the G.P. $\,\{\sqrt2, \sqrt6,3\sqrt2,3\sqrt6,\cdots\}\,$ is $\,243\sqrt2\,?$

Sol. Let the first term of the G.P. be $\,(a)=\sqrt2\,$ and the common ratio being $\,(r)=\frac{\sqrt6}{\sqrt2}=\sqrt3.$

Let the $\,n\,$th term of a G.P., denoted by $\,t_n\,$ and is given by 

$\,t_n=243\sqrt2\\ \Rightarrow \sqrt2(\sqrt3)^{n-1}=243\sqrt2\\ \Rightarrow (\sqrt3)^{n-1}=243=(\sqrt3)^{10}\\ \Rightarrow n-1=10\\ \Rightarrow n=10+1=11\rightarrow(1)$

Hence, from $\,(1),\,$ we can conclude that $\,11\,$ th term of the given G.P.is $\,243\sqrt2.$

$\,5(ii)\,$ Is $\,256\,$ a term of the G.P. $\,\{3,6,12,24,\cdots\}?$

Sol. Let the first term of the G.P. be $\,(a)=3\,$ and the common ratio being $\,(r)=\frac{6}{3}=2.$

Let the $\,n\,$th term of a G.P., denoted by $\,t_n\,$ and is given by

$\,t_n=ar^{n-1}\\ \Rightarrow 256=3 \cdot2^{n-1}\\ \Rightarrow 2^{n-1}=\frac{256}{3}\neq \text{an integer.}$

So, $\,256\,$ is not   a term of the G.P. $\,\{3,6,12,24,\cdots\}$.

$\,6.\,$ Fill in the gaps (indicated by _) in each of the following G.P.'s :

$\,(i)\,~~\frac{11}{4},-,-,\frac{22}{27},\cdots$

Sol. Let the first term of the G.P. be $\,(a)=\frac{11}{4}\,$ and the common ratio being $\,r.$

Let the $\,n\,$th term of a G.P., denoted by $\,t_n\,$ .

Now, by the given condition,

$\,t_4=\frac{22}{27}\\ \Rightarrow \frac{11}{4} \cdot r^{4-1}=\frac{22}{27}\\ \Rightarrow r^3=\frac{22}{27} \times \frac{4}{11}\\ \Rightarrow r^3=\frac{8}{27}=\left(\frac 23\right)^3\\ \Rightarrow r=\frac 23.$

So, $\,t_2=ar=\frac{11}{4}\cdot \frac 23=\frac{11}{6},\\~t_3=ar^2=\frac{11}{4} \cdot \left(\frac 23\right)^2=\frac{11}{9}.$

$\,6.\,$ Fill in the gaps (indicated by _) in each of the following G.P.'s :

$\,(iii)\,~~0.1,-,-,-,-,(0.000032),\cdots$

Sol. Let the first term of the G.P. be $\,(a)=0.1\,$ and the common ratio being $\,r.$

Let the $\,n\,$th term of a G.P., denoted by $\,t_n\,$ .

Now, by the given condition,

$\,t_6=0.000032\\ \Rightarrow (0.1)\cdot r^{6-1} =0.000032\\ \Rightarrow r^5=0.00032 =(0.2)^5\\ \Rightarrow r=0.2$

So, taking $\,a=0.1,r=0.2$,

we get ,$\,t_2=ar=(0.1)\times (0.2)=0.02,\\~t_3=ar^2=(0.1)\times (0.2)^2=0.004,\\~t_4=ar^3=(0.1)\times (0.2)^3=0.0008,\\~t_5=ar^4=(0.1) \times (0.2)^4=0.00016.$

$\,7.\,$ Find $\,r\,$ if $\,\,(i)~3r+1,\,7r,10r+8\,\,$ are in G.P. ; $\,(ii)\,2r,4r+1,6r+2\,\,$ are in G.P.

$\,(i)\,$ Sol.  Since $\,\,~3r+1,\,7r,10r+8\,\,$ are in G.P.,

$\,\therefore \,\, (7r)^2=(3r+1)(10r+8)\\ \Rightarrow 49r^2=30r^2+24r+10r+8 \\ \Rightarrow 19r^2-34r-8=0 \\ \Rightarrow 19r^2-38r+4r-8=0 \\ \Rightarrow 19r(r-2)+4(r-2)=0 \\ \Rightarrow (r-2)(19r+4)=0 \\ \Rightarrow r=2,~~\frac{-4}{19}.$

$\,(ii)\,$ Sol.  Since $\,\,2r,4r+1,6r+2\,\,$ are in G.P.,

$\,\therefore \,\, (4r+1)^2=(2r)(6r+2)\\ \Rightarrow 16r^2+8r+1=12r^2+4r \\ \Rightarrow 4r^2+4r+1=0 \\ \Rightarrow (2r)^2+2 \cdot (2r) \cdot 1+1^2=0 \\ \Rightarrow (2r+1)^2=0\\ \Rightarrow 2r+1=0 \\ \Rightarrow r=-\frac 12.$

$\,8.\,$ Insert four geometric means between $\,\frac 74\,$ and $\,56.$

Sol. Let $\,x,y,z,w\,$ be four geometric means between $\,\frac 74\,$ and $\,56,\,\,$ so that $\,\frac 74,x,y,z,w,56\,$ form G.P.

Here, $\,t_6=56 \\ \Rightarrow \frac 74 \cdot r^{6-1}=56 \\ \Rightarrow r^5=56 \times \frac 47\\ \Rightarrow r^5=32=2^5\\ \Rightarrow r=2.$

So, $\,t_2=x=ar=\frac 74 \cdot 2=\frac 72,\\~t_3=y=ar^2=\frac 74\cdot 2^2=7,\\~t_4=z=ar^3=\frac 74 \cdot 2^3=14,\\~t_5=w=ar^4=\frac 74 \cdot 2^4=28.$

$\,9.\,$ If three unequal positive numbers $\,a,b,c\,\,$ are in G.P. then show that, $\,a+c>2b.$

Sol. If three unequal positive numbers $\,a,b,c\,\,$ are in G.P., 

$\,\,\frac ba=\frac cb \\ \Rightarrow b^2=ac\rightarrow(1)$

Now, for three unequal positive numbers $\,a,b,c\,\,$ ,

$\,a+c-2\sqrt{ac}\\=(\sqrt a)^2+(\sqrt c)^2-2\sqrt a.\sqrt c\\=(\sqrt a-\sqrt c)^2 >0\\ \Rightarrow a+c-2\sqrt{ac}>0\\ \Rightarrow a+c-2\sqrt{b^2}>0\,\,\text{[By (1)]}\\ \Rightarrow a+c-2b>0 \\ \Rightarrow a+c>2b\,\,\text{(showed)}$

$\,10.\,$ Three different numbers $\,a,b,c\,$ are both in A.P. and G.P.-Is it possible ? Give reasons in support of your answer.

Sol. If three different numbers $\,a,b,c\,$ are in A.P.,

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

Again, if three different numbers $\,a,b,c\,$ are in G.P.,

$\,\frac ba=\frac cb \\ \Rightarrow b^2=ac\rightarrow(2)$

Now, from $\,(1),(2)\,\,$ we get,

$\,a+c=2b=2\sqrt{ac}\\ \Rightarrow (\sqrt a)^2-2\sqrt a.\sqrt c+(\sqrt c)^2=0\\ \Rightarrow (\sqrt a-\sqrt c)^2=0 \\ \Rightarrow \sqrt a=\sqrt c\\ \Rightarrow a=c\,~~~\text{which is not possible.}$

Hence, three different numbers $\,a,b,c\,$ can not be both in A.P. and G.P.