Differentiation (Part-38) | S N De

 

$\,17.\,$ Find the sum to $\,n\,$ terms :

$\,(i)\,\,\frac 12+\frac{3}{2^2}+\frac{5}{2^3}+\cdots +\frac{2n-1}{2^n}$

Sol. Let  $ \,S=\,\,\,\frac 12+\frac{3}{2^2}+\frac{5}{2^3}+\cdots +\frac{2n-1}{2^n}$ 

Now, $\,\,\frac 12 S=S-\frac 12S \\ \Rightarrow \frac 12S=\left(\frac 12+\frac{3}{2^2}+\frac{5}{2^3}+\cdots +\frac{2n-1}{2^n}\right)\\~~~~~~~~~~~~~-\frac 12 \left(\frac 12+\frac{3}{2^2}+\frac{5}{2^3}+\cdots +\frac{2n-1}{2^n}\right) \\ \Rightarrow \frac 12 S=\frac 12+\left[\left(\frac{3}{2^2}-\frac{1}{2^2}\right)+\left(\frac{5}{2^3}-\frac{3}{2^3}\right)\\+\cdots\text{to (n-1) terms}\right]-\frac{2n-1}{2 \cdot 2^n}\\ \Rightarrow \frac 12 S=\frac 12+\left[\frac 12+\frac{1}{2^2}\\+\cdots \text{to (n-1) terms}\right]-\frac{2n-1}{2 \cdot 2^n} \\ \Rightarrow \frac 12 S=\frac 12+\frac 12\cdot\frac{1-\frac{1}{2^{n-1}}}{1-\frac 12}-\frac{2n-1}{2 \cdot 2^n} \\ \Rightarrow \frac 12 S=\frac 12+\left(1-\frac{1}{2^{n-1}}\right)-\frac{2n-1}{2 \cdot 2^n} \\ \Rightarrow S=1+2 \left(1-\frac{1}{2^{n-1}}\right) -\frac{2n-1}{2^n}\\ \Rightarrow S=1+2-2\cdot 2^{-(n-1)}-(2n-1)2^{-n} \\ \Rightarrow S=3-2\cdot2^{-n}\cdot2-(2n-1)2^{-n}\\ \Rightarrow S=3-4 \cdot2^{-n}-2n \cdot2^{-n}+2^{-n} \\ \Rightarrow S=3-3\cdot2^{-n}-2n \cdot 2^{-n} \\ \Rightarrow S=3-(2n+1)2^{-n}\,\,\text{(ans.)}$

$\,17.\,$ Find the sum to $\,n\,$ terms :

$\,(ii)\,\,2+3 \cdot 3+4 \cdot 3^2+5 \cdot 3^3+\cdots$

Sol.  Let $\,\,S=2+3 \cdot 3+4 \cdot 3^2+5 \cdot 3^3+\cdots \text{to n terms} \\ \Rightarrow S=2+3 \cdot 3+4 \cdot 3^2+5 \cdot 3^3\\+\cdots+(n+1)3^{n-1} \rightarrow(1)\\ \Rightarrow 3S=~~~~~2 \cdot 3+3\cdot 3^2+4 \cdot 3^3+5 \cdot 3^4\\+\cdots+n \cdot 3^{n-1}+(n+1)3^n \rightarrow(2)$

Hence, from $\,(1)\,$ and $\,(2)\,\,$ we get,

$\,\,S-3S=2+[(3 \cdot 3-2 \cdot3)+(4 \cdot 3^2-3 \cdot 3^2)\\+(5 \cdot 3^3-4 \cdot3^3)+\cdots\text{to (n-1) terms}]\\-(n+1)3^n\\ \Rightarrow -2S=2+[3+3^2+3^3\\+\cdots \text{to (n-1) terms}]-(n+1)3^n \\ \Rightarrow -2S=2+3\cdot\frac{3^{n-1}-1}{3-1}-(n+1)3^n \\ \Rightarrow  -2S=2+\frac 32 \cdot (3^{n-1}-1)-(n+1)3^n \\ \Rightarrow  -2S= 2-\frac 32+\frac 32 \cdot \frac{3^n}{3}-(n+1)3^n \\ \Rightarrow  -2S=\frac12+\frac 12 \cdot 3^n-(n+1)3^n \\ \Rightarrow  -2S=\frac 12+3^n(\frac 12-n-1) \\ \Rightarrow  -2S=\frac 12-3^n \cdot (\frac 12+n) \\ \Rightarrow  -2S=\frac 12-\frac{3^n}{2}(2n+1) \\ \Rightarrow  S=-\frac 14+\frac{3^n}{4}(2n+1) \\ \Rightarrow  S=\frac 14[3^n(2n+1)-1]\,\,\,\text{(ans.)}$

$\,17.\,$ Find the sum to $\,n\,$ terms :

$\,(iii)\,\,12+105+1008+10011+\cdots $ 

Sol.  $\,\,S=12+105+1008+10011+\cdots \text{to n terms}\rightarrow(1)$

Let $\,r\,$th term of the series be denoted by $\,t_r.$

So, $\,\,\,t_r=10^r+3r-1 \\ \Rightarrow \sum_{r=1}^n t_r=\sum_{r=1}^n  10^r+3 \sum_{r=1}^n  r\\ ~~~~~~~~~~~~~~-\sum_{r=1}^n (1) \\ \Rightarrow S=10 \cdot\frac{10^n-1}{10-1}+3 \cdot \frac{n(n+1)}{2}-n \\ \Rightarrow S=\frac{10}{9}\cdot (10^n-1)+\frac{3n^2+3n-3n}{2} \\ \Rightarrow S=\frac{10}{9}\cdot(10^n-1)+\frac 12(3n^2+n)\,\,\text{(ans.)}$

$\,(iv)\,\,12+14+24+58+164+\cdots\text{to n terms}\\=2(6+7+12+29+82+\cdots\text{to n terms})\\=2[(1+5)+(3+4)+(9+3)+(27+2)\\~~~~~+(81+1)+\cdots\text{to n terms}] \rightarrow(1)$

Let $\,\,t_r\,$ be the $\,r\,$ th term of the series where 

$\,\,t_r=2(3^{r-1}+6-r) \\ \Rightarrow \sum_{r=1}^n t_r=2 \left[\sum_{r=1}^n 3^{r-1}+\sum_{r=1}^n 6\\~~~~~~~~~~~~~~~~~~~~~~-\sum_{r=1}^n (r)\right] \\ \Rightarrow S=2\left(\frac{3^n-1}{3-1}+6n-\frac{n(n+1)}{2}\right) \\ \Rightarrow S=3^n-1+12n-n^2-n\\ \Rightarrow S=3^n-n^2+11n-1\,\,\text{(ans.)}$

$\,18.\,$ If $\,r\,$th term of the series is $\,\,(2r+1)2^r;\,\,$ Find the sum of first $\,n\,$ terms of the series.

Sol.  If $\,r\,$th term of the series is denoted by $\,t_r,\,$ then $\,t_r=(2r+1)2^r ;$

Let $\,S=\sum_{r=1}^n t_r \\=\sum_{r=1}^n[(2r+1)2^r]\\=3 \cdot 2+5 \cdot2^2+\cdots+(2n-1)2^{n-1}\\+(2n+1)2^n\rightarrow(1) \\ \Rightarrow 2S=3 \cdot2^2+5 \cdot 2^3+\cdots\\+(2n-1)2^n+(2n+1)2^{n+1}\rightarrow(2)$

Hence, from $\,(1)\,$ and $\,(2)\,$ we get,

$\,2S-S=(3-5)\cdot 2^2+(5-7)\cdot  2^3\\+\cdots+\{(2n-1)-(2n+1)\}\cdot 2^n+(2n+1)2^{n+1}-3 \cdot 2^1 \\ \Rightarrow S=-2\cdot  2^2-2 \cdot  2^3-\cdots-2\cdot  2^n\\+(2n+1)2^{n+1}-3 \cdot  2 \\ \Rightarrow S=-2 \cdot  2-2\cdot  2^2-2 \cdot  2^3-\cdots\\-2\cdot  2^n+(2n+1)2^{n+1} -1\cdot  2 \\ \Rightarrow S=-2[2+2^2+2^3+\cdots+2^n]\\+(2n+1)2^{n+1}-2 \\ \Rightarrow S=-2 \cdot \left[2 \cdot \frac{2^n-1}{2-1}\right]+(2n+1)2^{n+1}-2 \\ \Rightarrow S=-4(2^n-1)+(4n+2)2^{n}-2 \\ \Rightarrow S=2^n(4n+2-4)+4-2 \\ \Rightarrow S=(4n-2)2^n+2 \\ \therefore S=(2n-1)2^{n+1}+2\,\,\text{(ans.)}$

$\,19.\,$ If one geometric mean $\,G\,$ and two arithmetic means $\,p,\,q\,$ be inserted between two given numbers, then prove that,

$\,\,G^2=(2p-q)(2q-p.)$

Sol. Let the two numbers be $\,a\,$ and $\,b\,$ . 

Then, $\,\,G= \sqrt{ab} \Rightarrow G^2=ab\rightarrow(1)$

Also, $\,p\,$ and $\,q\,$ are two A.M.'s between $\,a\,$ and $\,b\,$.

$\,\therefore \,a,p,q,b\,\,$ are in A.P.

$\,\therefore \,\,p−a=q−p\,\,\,\text{and}\,\, q−p=b−q$

So, $\,a=2p−q\,$ and $\,\,b=2q−p\rightarrow(2)$

Hence, from $\,(1)\,$ and $\,(2),\,$ we get,

$\,\,G^2=(2p-q)(2q-p)\,\,\text{(proved)}$

$\,20.\,$  $\,\,A,B,C\,$  have together Rs. $\,5700\,$ and the amount of money possessed by them form a G.P. If $\,B\,$ had Rs. $\,150\,$ more , the amounts would form an A.P. Find the amount they possess.

Sol. Let $\,A,B,C\,$ possess $\,x,y,z\,$ amount of money respectively. 

So, $\,\,x+y+z=5700 \rightarrow(1)$

Also, by the given condition, $\,\,y^2=zx \rightarrow(2)$

If $\,B\,$ had Rs. $\,150\,$ more , the amounts would form an A.P. and this means 

$\,\,2(y+150)=x+z\rightarrow(3)$

Hence, from $\,(1)\,$ and $\,(3),\,$ we get,

$\,y+2(y+150)=5700 \\ \Rightarrow 3y+300=5700 \\ \Rightarrow 3y=5700-300 \\ \Rightarrow y=\frac{5400}{3}=1800.$

Again, by $\,(2),\,$ we get

$\,(1800)^2=zx \\ \Rightarrow z=\frac{3240000}{x}\rightarrow(4)$

Now, from $\,(1)\,$ and $\,(4),\,$ we get,

$\,x+z=5700-1800 \\ \Rightarrow x+\frac{3240000}{x}=3900 \\ \Rightarrow x^2+3240000=3900x \\ \Rightarrow x^2-3900x+3240000=0 \\ \Rightarrow x^2-1200x-2700x+3240000=0 \\ \Rightarrow x(x-1200)-2700(x-1200)=0 \\ \Rightarrow (x-1200)(x-2700)=0 \\ \therefore x=1200, 2700.$

If $\,x=1200,\,\,z=\frac{3240000}{1200}=2700;$

and if $\,x=2700,\,\,z=\frac{3240000}{2700}=1200;$ 

The amount they possess are $\,\,1200, 1800, 2700\,\,$ or $\,\, 2700,1800, 1200,\,\,$ respectively. 

$\,21.\,$ If the $\,(m+1)\,$-th , $\,(n+1)\,$th  and $\,(r+1)\,$ th terms of an A.P. are in G.P. and $\,\frac 1m,\frac 1n,\frac 1r\,$ are in A.P. , then find the ratio of the first term of the A.P. to its common difference in terms of $\,n\,.$

Sol.  Let $\,a\,$ be the first term and $\,d\,$ be the common difference of the A.P. Also suppose that $\,t_n\,\,$ denotes the $\,n\,$ th term of the A.P.

So, $\,t_{m+1}=a+(m+1-1)d=a+md, \\~~ t_{n+1}=a+nd,\\~~t_{r+1}=a+rd.$ 

Now, since $\,\,t_{m+1},\,t_{n+1},\,t_{r+1}\,\,$ are in G.P. ,

$\,t_{n+1}^2=t_{m+1} \cdot t_{r+1} \\ \Rightarrow (a+nd)^2=(a+md)\times (a+rd) \\ \Rightarrow  a^2+2and +n^2d^2=a^2+ard\\~~~~~~~~~~~+amd+mrd^2 \\ \Rightarrow  ad(2n-r-m)=d^2(mr-n^2) \\ \Rightarrow \frac ad=\frac{mr-n^2}{2n-r-m} \rightarrow(1)$

Since, $\,\frac 1m,\frac 1n,\frac 1r\,$ are in A.P. 

$\,\frac 2n=\frac 1m+\frac 1r \\ \Rightarrow  \frac 2n=\frac{r+m}{mr} \\ \Rightarrow  n=\frac{2mr}{r+m}\rightarrow(2).$

Hence , from $\,(1)\,$ and $\,(2),\,\,$ we get,

$\,\,\frac ad=\frac{mr-\left(\frac{2mr}{r+m}\right)^2}{\frac{4mr}{r+m}-r-m}\\~~~~~=\frac{mr\left[1-\frac{4mr}{(r+m)^2}\right]}{\frac{4mr}{r+m}-(r+m)}\\~~~~~=\frac{mr}{(r+m)} \times \frac{(r+m)^2-4mr}{4mr-(r+m)^2}\\~~~~~=\frac n2 \times (-1) \cdot\frac{4mr-(r+m)^2}{4mr-(r+m)^2}\,\,~~[\text{By (2)}]\\~~~~~=-\frac n2\,\,\text{(ans.)}$

$\,22.\,$  The sum of first ten terms of an A.P. is $\,155\,$ and the sum of first two terms of a G.P. is $\,9\,$. The first term of the A.P. is equal to the common ratio of the G.P. and the first term of the G.P. is equal to the common difference of the A.P. Find the two progressions.

Sol. Let the A.P.  be given by : $\,\,a,a+d,a=2d,\cdots $ and the G.P.  be given by $\,\,d,da,da^2, \cdots$

Now, by the given condition, we have,

$\,S_{10}=155 \\ \Rightarrow \frac{10}{2}[2a+(10-1)d]=155  \\ \Rightarrow 5(2a+9d)=155 \\ \Rightarrow 2a+9d=\frac{155}{5}=31 \\ \therefore a=\frac{31-9d}{2}\rightarrow (1)$

Again, $\,d+da=9 \Rightarrow d(1+a)=9 \rightarrow(2)$

Hence, from $\,(1)\,$ and $\,(2)\,$, we get,

$\,\,d \left[1+\frac 12(31-9d)\right]=9 \\ \Rightarrow d(2+31-9d)=9 \times 2 \\ \Rightarrow d(33-9d)=9 \times 2 \\ \Rightarrow 3d(11-3d)=18 \\ \Rightarrow d(11-3d)=6 \\ \Rightarrow 3d^2-11d+6=0 \\ \Rightarrow 3d^2-9d-2d+6=0 \\ \Rightarrow 3d(d-3)-2(d-3)=0 \\ \Rightarrow (d-3)(3d-2)=0 \\ \Rightarrow d=3, \frac 23.$

For $\,\,d=3,\,$ we get from $\,(2),\,$

$\,\,3(1+a)=9 \Rightarrow a=2$ 

So, the A.P. is : $\,\,2,\,5,\,8, \cdots $ and the G.P. is : $\,\,3,\,6,\,12 \cdots$

For $\,\,d=\frac 23,\,$ we get from $\,(2),\,$

$\,\,\frac 23(1+a)=9 \Rightarrow a=\frac{25}{2}$ 

So, in this case, the A.P. is : $\,\,\frac{25}{2},\,\frac{25}{2}+\frac 23,\,\frac{25}{2}+2 \cdot \frac 23, \cdots $ 

and the G.P. is : $\,\,\frac 23,\,\frac 23 \cdot \frac{25}{2},\,\frac 23 \cdot \left(\frac{25}{2}\right)^2, \cdots$