Differentiation (Part-38) | S N De

 

$\,1.\,$ The common ratio of a series in G.P. is $\,3;\,$ the sum of the $\,1\,$st and $\,3\,$rd terms is equal to the sum of the squares of the $\,1\,$st and $\,2\,$nd terms; find the sum of $\,n\,$ terms. If $\,n = 6,\,$ show that the sum is $\,\,364.$

Sol. Let the first term of G.P. be $\,a.$  So, the G.P. can be written as : $\,a,\,3a,\,3^2a,\cdots.$

Since  the sum of the $\,1\,$st and $\,3\,$rd terms is equal to the sum of the squares of the $\,1\,$st and $\,2\,$nd terms ;

$\,t_1+t_3=t_1^2+t_2^2 \\ \Rightarrow a+a \cdot 3^{3-1}=a^2+(a \cdot3^{2-1})^2 \\ \Rightarrow a+9a=a^2+a^2 \cdot9 \\ \Rightarrow 10a=10a^2 \\ \Rightarrow  a=1 \,\,\,[\because a \neq 0] $

So,  the sum of $\,n\,$ terms $\,S_n=a \cdot \frac{3^n-1}{3-1}\\~~~~~=1 \cdot \frac{3^n-1}{2}\\~~~~~=\frac 12(3^n-1)\,\,\,\text{(ans.)}$

For $\,n=6,\,\,S_6=\frac 12(3^6-1)=\frac 12(729-1)=368\,\,\text{(showed)}$

$\,2.\,$ If the fourth term of a series in geometric progression is $\,24\,$ and the seventh term is $\,192\,$, find the sum of its first ten terms.

Sol. Let the first term of G.P. be $\,a\,$, common ratio $=r,\,$ the $\,n\,$ th term being $\,t_n.$

$\,t_4=a \cdot r^{4-1}=a\cdot r^3=24 \rightarrow(1) \\ t_5=a \cdot r^{7-1}= a \cdot r^6=192 \rightarrow(2).$

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

$\,\,\frac{ar^6}{ar^3}=\frac{192}{24} \\ \Rightarrow r^3=8 \\ \Rightarrow r=\sqrt[3]{8}=2.$

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

$\,a\cdot 2^3=24 \Rightarrow a=\frac{24}{8}=3.$

So, $\,\,S_{10}=3 \cdot \frac{2^{10}-1}{2-1}\\~~~~~~~=3 \cdot (1024-1)\\~~~~~~~=3 \times 1023\\~~~~~~~=3069\,\,\text{(ans.)}$ 

$\,3.\,$ How many terms of the series $\,64+32+16+8+\cdots$ must be taken so that the sum may be $\,\,127\frac 12?$  

Sol. In the given series, the first term $\,(a)=64,\,$ the common ratio $\,(r)=\frac{32}{64}=\frac 12.$

So, $\,S_n=a\cdot\frac{1-r^n}{1-r} \\ \Rightarrow 127\frac 12=64 \cdot\frac{1-(1/2)^n}{1-1/2}\\ \Rightarrow \frac{255}{2}=(64 \times 2)\cdot[1-(\frac 12)^n] \\ \Rightarrow 1-(\frac 12)^n=\frac{255}{128 \times 2} \\ \Rightarrow 1-\frac{255}{256}=(\frac 12)^n \\ \Rightarrow (\frac 12)^n=\frac{1}{256} \\ \Rightarrow(\frac 12)^n=(\frac 12)^8 \\ \Rightarrow n=8.$

Hence, $\,8\,$ terms of the given series must be taken so that the sum may be $\,\,127\frac 12.$

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

$\,4(i).\,$ A G.P. has first term $\,3\,$ and the last term $\,48.\,$ If each term is twice the preceding , find the number of terms and the sum of the G.P.

Sol. A G.P. has first term $(a)=\,3\,$ and the last term $(t_n)=\,48.\,$ 

Second term $\,(t_2)=2 \times 3=6,\,\,$ so the common ratio is : $\,(r)=\frac{6}{3}=2.$

So,$\,\,\,t_n=a \cdot r^{n-1} \\ \Rightarrow 48=3 \cdot 2^{n-1} \\ \Rightarrow \frac{48}{3}=2^{n-1}  \\ \Rightarrow 16=2^{n-1}  \\ \Rightarrow 2^{n-1}=2^4  \\ \Rightarrow n-1=4  \\ \Rightarrow n=4+1=5.$

Hence, the number of terms of the given G.P. is $\,5.$

Now, the sum of G.P. is : $\,S_5=3 \cdot \frac{2^5-1}{2-1}\\~~~~~=3(2^5-1)\\~~~~~=3(32-1)\\~~~~~=3 \times 31 \\~~~~~= 93\,\,\text{(ans.)}$

$\,4(ii)\,\,$ The sum of first $\,8\,$ terms of a G.P. is $\,5\,$ times the sum of the first $\,4\,$ terms. Find the common ratio.

Sol.  Let sum of $\,n\,$ terms of the given G.P. be denoted by $(S_n)\,$ and common ratio being $\,\,\,r,\,\,$ and the first term be denoted by $\,a.$

$\,\,S_8=5 \times S_4 \\ \Rightarrow a \cdot \frac{r^8-1}{r-1}=5 \times a \cdot \frac{r^4-1}{r-1} \\ \Rightarrow r^8-1=5(r^4-1)\\ \Rightarrow (r^4+1)(r^4-1)=5(r^4-1)\\ \Rightarrow (r^4-1)(r^4+1-5)=0 \rightarrow (1) $

So, from $\,(1),\,$ we get,

$\,r^4-1=0 \\ \Rightarrow r^4=1 \\ \Rightarrow r=\pm 1.$

Again, by $\,(1),\,$ we get, 

$\,r^4+1-5=0\\ \Rightarrow r^4=4 \\ \Rightarrow r^2=2 \\ \Rightarrow r= \pm \sqrt{2}$

$\,4(iii)\,\,$ The sum of first four terms of a G.P. is $\,40\,$ and the sum of first $\,8\,$ terms is $\,\,3280.\,$ Obtain the G.P. 

Sol. The sum of first $\,n\,$ terms of a G.P. is :

$\,S_n=a \cdot \frac{r^n-1}{r-1},\,$ where $\,r\,$ is the common ratio, $\,a\,$ being the first term of the G.P.

So, $\,\,S_4=a\cdot \frac{r^4-1}{r-1} \\ \Rightarrow 40=a \cdot \frac{r^4-1}{r-1}\rightarrow(1)$

Again, $\,\,S_8=a \cdot \frac{r^8-1}{r-1} \\ \Rightarrow 3280=a \cdot \frac{r^8-1}{r-1} \rightarrow(2)$

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

$\,\,\frac{3280}{40}=\frac{a \cdot \frac{r^8-1}{r-1}}{a \cdot \frac{r^4-1}{r-1}}\\ \Rightarrow 82=\frac{r^8-1}{r^4-1} \\ \Rightarrow 82=\frac{x^2-1}{x-1},\,\,[\text{where}~~~\,x=r^4] \\ \Rightarrow 82(x-1)=x^2-1 \\ \Rightarrow x^2-82x-1+82=0 \\ \Rightarrow x^2-82x+81=0 \\ \Rightarrow x^2-x-81x+81=0 \\ \Rightarrow x(x-1)-81(x-1)=0 \\ \Rightarrow (x-1)(x-81)=0 \\ \Rightarrow x=1,81.$

For $\,x=1 \\ \Rightarrow r^4=1 \\ \Rightarrow r =\pm 1.$

For $\,r=\pm 1,\,\,$ the series becomes $\,\,a,a,a,\cdots $ or $\,\,-a,-a,-a,\cdots $ which is not possible.

For $\,\,x=81 \\ \Rightarrow r^4=81 \\ \Rightarrow r^2=9 \\ \Rightarrow r =\pm 3.$

Now, for $\,r=3,\,$ we get from $\,(1),\,$

$\,40=a \cdot \frac{3^4-1}{3-1} \\ \Rightarrow 40=a \cdot\frac{81-1}{2} \\ \Rightarrow 40=a \cdot\frac{80}{2}\\ \Rightarrow 40=40a \\ \Rightarrow a=1.$

So, for $\,r=3,\,\,\,a=1\,$ the series becomes :

$\,a, ar, ar^2,ar^3, \cdots \\=1,3,9,27,\cdots\,\,\text{(ans.)}$

Similarly, for $\,r=-3,\,$ we get from $\,(1),\,$

$\,40=a \cdot \frac{(-3)^4-1}{-3-1} \\ \Rightarrow 40=a \cdot\frac{81-1}{-4} \\ \Rightarrow 40=a \cdot\frac{80}{-4}\\ \Rightarrow 40=-20a \\ \Rightarrow a=-2.$

So, for $\,r=3,\,\,\,a=1\,$ the series becomes :

$\,a, ar, ar^2,ar^3, \cdots \\=-2,6,-18,54,\cdots\,\,\text{(ans.)}$

$\,4(iv)\,\,$ The first and last terms of a G.P. having finite number of terms are respectively $\,3\,$ and $\,768\,$ and the sum of the G.P. is $\,1533.\,$  Find the number of terms in the G.P. and its common ratio.

Sol. Let sum of $\,n\,$ terms of the given G.P. be denoted by $(S_n)\,$ and common ratio being $\,\,\,r,\,\,$ and the first term be denoted by $\,a.$

Here, $\,a=3,\,t_n=768,\,S_n=1533.$

$\,\,t_n=768 \\ \Rightarrow 3\cdot r^{n-1}=768 \\ \Rightarrow r^{n-1}=\frac{768}{3} \\ \Rightarrow r^{n-1}=256 =2^8 \\ \Rightarrow n-1=8 \\ \Rightarrow n=8+1=9.$

Again, $\,\,S_n=1533 \\ \Rightarrow 3 \cdot \frac{r^n-1}{r-1}=1533 \\ \Rightarrow \frac{r^n-1}{r-1}=\frac{1533}{3} \\ \Rightarrow r^n-1=511(r-1) \\ \Rightarrow r^n-511r=-511+1 \\ \Rightarrow r(r^{n-1}-511)=-510 \\ \Rightarrow r(256-511)=-510 \\ \Rightarrow r \times (-255)=-510 \\ \Rightarrow r=\frac{-510}{-255}=2.$

Hence, the number of terms in the G.P. is $\,9\,$ and its common ratio is $\,2.$