Differentiation (Part-38) | S N De

 

$\,11.\,$ If $\,a,b,c\,$ form an A.P. and $\,b,c,a\,\,$ form a G.P. , show that $\,\frac 1c,\frac 1a,\frac 1b\,$ form an A.P.

Sol. Since $\,a,b,c\,$ form an A.P. , 

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

Again, since $\,b,c,a\,\,$ form a G.P., 

$\,\frac cb=\frac ac \Rightarrow c^2=ab \rightarrow(2)$

From $\,(1),\,$ we get,

$\,2b=a+c \\ \Rightarrow \frac{2b}{ab}=\frac{a}{ab}+\frac{c}{ab} \\ \Rightarrow \frac 2a=\frac 1b+\frac{c}{c^2}\,\,\,\,[\text{By (2)}]\\ \Rightarrow \frac 2a=\frac 1b+\frac 1c\rightarrow(3)$

Hence, from $\,(3),\,$ we can conclude that $\,\frac 1c,\frac 1a,\frac 1b\,$ form an A.P.

$\,12.\,$ If the arithmetic mean of $\,y\,$ and $\,z\,$ be $\,x\,$ and that of $\,\frac 1x\,$ and $\,\frac 1y\,$ be $\,\frac 1z\,$, then show that $\,x,y,z\,$ are in G.P.

Sol.  If the arithmetic mean of $\,y\,$ and $\,z\,$ be $\,x\,$, then

$\,x=\frac{y+z}{2} \rightarrow(1)$

Again, if the arithmetic mean of $\,\frac 1x\,$ and $\,\frac 1y\,$ be $\,\frac 1z\,$,then

$\,\frac 1z=\frac 12\left(\frac 1x+\frac 1y\right) \\ \Rightarrow \frac 2z=\frac 1x+\frac 1y \\ \Rightarrow  \frac 2z-\frac 1x=\frac 1y \\ \Rightarrow \frac 1y=\frac{2x-z}{zx} \\ \Rightarrow  \frac{zx}{y}=2x-z \\ \Rightarrow \frac{zx}{y}=(y+z)-z\,\,\,[\text{By (1)}] \\ \Rightarrow  \frac{zx}{y}=y \\ \Rightarrow  zx=y^2 \\ \Rightarrow  \frac yx= \frac zy\rightarrow(2)$

Hence, by $\,(2),\,$ we can conclude that $\,x,y,z\,$ are in G.P.

$\,13(i).\,$ The sum of three numbers in A.P. is $\,15.\,\,$ If $\,1,4\,$ and $\,19\,$ are added to the numbers the resulting numbers are in G.P. Find the numbers. 

Sol. Let $\,a-d,a,a+d\,$ are the three numbers in AP.

Now, given sum $= 15.$

$\therefore \,\, a-d+a+a+d = 15 \\ \Rightarrow 3a = 15 \\ \Rightarrow a = 15/3 = 5$

When added $\,1,4,19\,$ respectively they are in G.P.

$\, \therefore \, a-d+1, a+4, a+d+19\,\,$ are in G.P.

So $\, (a+4)^2 = (a-d+1) (a+d+19) \rightarrow (1)$

Substitute $\,a = 5,$ we get from $\,(1),$

$\,92 = (6-d)(24+d) \\ \Rightarrow 81 = 144-24d+6d-d^2 \\ \Rightarrow  d^2+18d-63 = 0 \\ \Rightarrow (d+21)(d-3) = 0 \\ \Rightarrow  d = -21 \,\, \text{or}\,\, 3 .$

So the numbers are $\,26, 5, -16\,$  or  $\,2, 5, 8.$

$\,13(ii)\,$ The sum of three numbers in A.P. is $\,18.$ If $\,2,2,6\,$ are added respectively to $\,1\,$st , $\,2\,$nd and $\,3\,$rd numbers , the resulting numbers are in G.P. Find the numbers.

Sol. Let $\,a-d,a,a+d\,$ are the three numbers in AP.

Now, given sum $= 18.$

$\therefore \,\, a-d+a+a+d = 18 \\ \Rightarrow 3a = 18 \\ \Rightarrow a = 18/3 = 6$

When added $\,2,2,6\,$ respectively they are in G.P.

$\, \therefore \, a-d+2, a+2, a+d+6\,\,$ are in G.P.

So $\, (a+2)^2 = (a-d+2) (a+d+6) \rightarrow (1)$

Substitute $\,a = 6,$ we get from $\,(1),$

$\,(6+2)^2 = (6-d+2)(6+d+6) \\ \Rightarrow 64=(8-d)(12+d) \\ \Rightarrow 64 = 96-12d+8d-d^2 \\ \Rightarrow  d^2+4d-32 = 0 \\ \Rightarrow d^2+8d-4d-32 = 0 \\ \Rightarrow  d(d+8)-4(d+8)=0 \\ \Rightarrow (d+8)(d-4)=0 \\ \Rightarrow d=-8\,\, \text{or}\,\, 4 .$

So the numbers are $\,2, 6, 10\,$  or $\,14,6,-2.$

$\,13(iii)\,\,$ Three terms are in G.P. Their product is $\,216.\,$ If $\,4\,$ be added to the first term and $\,6\,$ to the second term, the resulting numbers are in A.P. Obtain the terms in G.P. 

Sol. Let the three numbers be $\,\frac ar,a,ar;\,\,\text{where r being common ratio.}$

So, $\,\frac ar\cdot a\cdot (ar)=216 \\ \Rightarrow a^3=216 \\ \Rightarrow a=\sqrt[3]{216}=6.$

Again, if $\,4\,$ be added to the first term and $\,6\,$ to the second term, the resulting numbers are in A.P. 

So, $\,2(a+6)=(\frac ar+4)+ar \\ \Rightarrow 2(6+6)=\frac 6r+4+6r\,\,[\because a=6]\\ \Rightarrow 2\times 12-4=6(r+\frac 1r) \\ \Rightarrow \frac{20}{6}=r+\frac 1r \\ \Rightarrow  \frac{10}{3}=\frac{r^2+1}{r} \\ \Rightarrow 10r=3r^2+3 \\ \Rightarrow 3r^2-10r+3=0 \\ \Rightarrow 3r^2-9r-r+3=0 \\ \Rightarrow 3r(r-3)-1(r-3)=0 \\ \Rightarrow (r-3)(3r-1)=0 \\ \Rightarrow r=3,\frac 13.$

For $\,r=3,\,$ three numbers are :

$\,\frac ar, a, ar \\ \Rightarrow \frac 63, 6, 6\times 3 \\ \Rightarrow 2,6,18\,\text{(ans.)}$

Again, for $\,r=\frac 13,\,$ three numbers are :

$\,\frac ar, a, ar \\ \Rightarrow \frac {6}{\frac 13}, 6, 6\times \frac 13 \\ \Rightarrow 18,6,2\,\text{(ans.)}$

$\,14.\,$ Find the sum to $\,n\,$ terms of each of the following series :

$\,(i)\,1+11+111+\cdots $

Sol. $\,1+11+111+\cdots \text{to n terms }\\=\frac 19(9+99+999+\cdots \text{to n terms })\\=\frac 19[(10-1)+(10^2-1)+(10^3-1)+\cdots \text{to n terms }]\\=\frac 19[(10+10^2+10^3+\cdots \text{to n terms })-n]\\=\frac 19 \times 10 \cdot\frac{10^n-1}{10-1}-\frac n9.\\=\frac{10}{81}\cdot(10^n-1)-\frac n9\,\,\text{(ans.)}$

$\,14.\,$ Find the sum to $\,n\,$ terms of each of the following series :

$\,(ii)\,2+22+222+\cdots $

Sol. $\,2+22+222+\cdots \text{to n terms }\\=\frac 29(9+99+999+\cdots \text{to n terms })\\=\frac 29[(10-1)+(10^2-1)+(10^3-1)+\cdots \text{to n terms }]\\=\frac 29[(10+10^2+10^3+\cdots \text{to n terms })-n]\\=\frac 29 \times 10 \cdot\frac{10^n-1}{10-1}-\frac {2n}{9}.\\=\frac{20}{81}\cdot(10^n-1)-\frac {2n}{9}\,\,\text{(ans.)}$

$\,14.\,$ Find the sum to $\,n\,$ terms of each of the following series :

$\,(iii)\,0.8+0.88+0.888+\cdots $

Sol. $\,0.8+0.88+0.888+\cdots \text{to n terms }\\=8(0.1+0.11+0.111+\cdots \text{to n terms })\\=\frac 29(0.9+0.99+0.999+\cdots \text{to n terms })\\=\frac 89[(1-0.1)+(1-0.01)+(1-0.001)+\cdots \text{to n terms }]\\=\frac 89[n-(0.1+0.01+0.001+\cdots \text{to n terms })]\\=\frac 89 \times n-\frac 89\cdot\frac{1}{10}\cdot\frac{1-(1/10)^n}{1-\frac{1}{10}}\\=\frac {8n}{9}-\frac 89\cdot \frac{1}{10}\cdot \frac{1-10^{-n}}{\frac{9}{10}}\\=\frac{8n}{9}-\frac{8}{81}\cdot(1-10^{-n})\,\,\text{(ans.)}$

$\,14.\,$ Find the sum to $\,n\,$ terms of each of the following series :

$\,(iv)\,0.6+0.66+0.666+\cdots $

Sol. $\,0.6+0.66+0.666+\cdots \text{to n terms }\\=6(0.1+0.11+0.111+\cdots \text{to n terms })\\=\frac 69(0.9+0.99+0.999+\cdots \text{to n terms })\\=\frac 23[(1-0.1)+(1-0.01)+(1-0.001)+\cdots \text{to n terms }]\\=\frac 23[n-(0.1+0.01+0.001+\cdots \text{to n terms })]\\=\frac 23 \times n-\frac 23\cdot\frac{1}{10}\cdot\frac{1-(1/10)^n}{1-\frac{1}{10}}\\=\frac {2n}{3}-\frac 23\cdot \frac{1}{10}\cdot \frac{1-10^{-n}}{\frac{9}{10}}\\=\frac{2}{3}\left[n-\frac{1}{9}\cdot(1-10^{-n})\right]\,\,\text{(ans.)}$

$\,14.\,$ Find the sum to $\,n\,$ terms of each of the following series :

$\,(v)\,1+4+13+40\cdots $

Sol. Let $\,S=1+4+13+40\cdots +\text{to n terms}$

Suppose $\,n\,$ th terms of $\,S\,$ be denoted by $\,t_n.$

So, $\,\,S=1+4+13+40+\cdots +t_n \\ ~S=~~~~~~~~1+4+13+\cdots+t_{n-1}+t_n\\-------------------\\ \text{Subtracting,}\,~0=1+3+3^2+\cdots\text{to n terms}-t_n \\ \Rightarrow t_n=1+3+3^2+\cdots \text{to n terms} \\ \Rightarrow t_n=1 \cdot\frac{3^n-1}{3-1} \\ \Rightarrow t_n=\frac 12(3^n-1) \rightarrow(1)$

Hence, by $\,(1),\,$ we get,

$\,t_1+t_2+t_3+\cdots+t_n\\=\frac 12[(3-1)+(3^2-1)+(3^3-1)+\cdots +(3^n-1)]\\=\frac 12[(3+3^2+3^3+\cdots +3^n)-n]\\=\frac 12 \cdot 3 \cdot \frac{3^n-1}{3-1}-\frac n2\\=\frac 34 (3^n-1)-\frac n2\,\,\,\text{(ans.)}$

$\,15(i)\,$ Three unequal numbers $\,a,b,c\,$ are in A.P. and $\,a,(b-a),(c-a)\,$ are in G.P. Prove that, $\, a:b:c=1:3:5.$

Sol.  Since $\,a,b,c\,$ are in A.P. , 

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

Again, since $\,a,(b-a),(c-a)\,$ are in G.P. , 

$\,(b-a)^2=a(c-a) \\ \Rightarrow  (b-a)^2=a(2b-a-a)\,\,[\text{By (1)}] \\ \Rightarrow (b-a)^2=a(2b-2a) \\ \Rightarrow  (b-a)^2=2a(b-a) \\ \Rightarrow  b-a=2a \,\,[\because b \neq a] \\ \Rightarrow b=2a+a=3a \rightarrow(2)$

So, from $\,(1),\,(2)\,$ we get, 

$\,c=2 \times 3a-a=5a.$

So, $\,a : b:c=a: 3a: 5a=1:3:5\,\text{(proved)}$

$\,15(ii)\,\,$ If $\,a,c,b\,$ are in A.P. and $\,b,c,d\,$ are in G.P. , prove that , $\,b,(b-c),(d-a)\,$ are in G.P.

Sol. If $\,a,c,b\,$ are in A.P., $~~~\,c-a=b-c \\ \Rightarrow a=2c-b \rightarrow(1)$

Again, since $\,b,c,d\,$ are in G.P., $\,~~\frac cb=\frac dc=r\,,\text{(say)}\\ \Rightarrow c=br, \\ ~~~~d=cr=(br)r=br^2 $

From $\,(1),\,$ we get, $\,a=2(br)-b=b(2r-1)$

Now, we have to prove that, $\,b,(b-c),(d-a)\,$ are in G.P.  and for that we have to prove that $\,\,\frac{b-c}{b}=\frac{d-a}{b-c}$

Now, $\,\frac{b-c}{b}\\=\frac{b-br}{b}\\=\frac{b(1-r)}{b}\\=1-r \rightarrow(2)$

And $\,\,\frac{d-a}{b-c}\\=\frac{br^2-b(2r-1)}{b-br}\\=\frac{b(r^2-2r+1)}{b(1-r)}\\=\frac{(1-r)^2}{1-r}\\=1-r \rightarrow(3)$

Hence, from $\,(2),(3)\,$ we can conclude that, $\,b,(b-c),(d-a)\,$ are in G.P.

$\,16.\,$ First term of an A.P. is the same as that of a G.P. the common difference of the one and the common ratio of the other are both $\,4.\,$  If the sum of the first three terms of each series are same , find the fourteenth term of each series.

Sol.  Let the first term of both series be $\,a\,$ . 

Now, suppose that the first three terms of A.P. are $\,a,a+4,a+8\,$ and that of G.P. are $\,\, a, 4a, 4^2a=16a.$

If the sum of the first three terms of each series are same , then

$\,a+(a+4)+(a+8)=a+4a+16a \\ \Rightarrow 3a+12=21a \\ \Rightarrow 12=21a-3a \\ \Rightarrow 18a=12 \\ \Rightarrow a=\frac{12}{18}=\frac 23.$

So, the fourteenth term of A.P. is :

$=a+(14-1)d\\=\frac 23+13 \times 4\\=\frac 23+52\\=52\frac 23\,\,\text{(ans.)}$

and  the fourteenth term of G.P. is :

$=a\cdot r^{14-1}\\=\frac 23 \cdot 4^{13}\\=\frac 23 \cdot 2^{26}\\=\frac 13 \cdot 2^{27}\,\,\text{(ans.)}$