Differentiation (Part-38) | S N De

 $\,5.\,$ The $\,n\,$th term of an arithmetic progression is $\,3n-1.\,$ Find the progression.

Sol. Let the $\,n\,$th term of an arithmetic progression be denoted by $\,t_n.$ 

So, $\,t_n=3n-1\,\,\text{(Given)}\\ t_1=3 \times 1-1=2,\\t_2=3 \times 2-1=5,\\t_3=3\times 3-1=8,\\ t_4=3\times 4-1=11, \\ \vdots $

Hence, the arithmetic progression is :

$\,\{2,5,8,11,\cdots\}.$

$\,6.\,$ Find the middle term (or terms) and the sum of the each of the following series :

$\,(i)\,~~~2+5+8+ \cdots+152.$

Sol. Here, the first term $\,(a)=2,\,$ the common difference $\,(d)=5-2=3\,$ and the last term $(l)=152.$

Let $\,t_n=152\\ \Rightarrow a+(n-1)d=152 \\ \Rightarrow 2+(n-1)\times 3=152 \\ \Rightarrow 3(n-1)=152-2 \\ \Rightarrow n-1=\frac{150}{3} \\ \Rightarrow n=50+1 \\ \Rightarrow n=51.$

So, the total number of terms of the series is : $\,51\,$ and the middle term is $\,\left(\frac{51+1}{2}\right)\,=26\,$ th term .

So, $\,t_{26}=2+(26-1)\times 3 \\ \Rightarrow t_{26}=2+25 \times 3=77\,\,\text{(ans.)}$ 

The sum of the series : 

$\,S=\frac n2(a+l) \\ \Rightarrow S=\frac{51}{2}(2+152) \\ \Rightarrow S=\frac{51}{2} \times 154 \\ \Rightarrow S=51 \times 77 \\ \Rightarrow S= 3927\,\text{(ans.)}$

$\,6.\,$ Find the middle term (or terms) and the sum of the each of the following series :

$\,(ii)\,~~~\frac 12+\frac 13+\frac 16+ \cdots +(-\frac 56)$

Sol. Here, the first term $\,(a)=\frac 12,\,$ the common difference $\,(d)=\frac 13-\frac 12=-\frac 16\,$ and the last term $(l)=-\frac 56.$

Let $\,t_n=-\frac 56\\ \Rightarrow a+(n-1)d=-\frac 56 \\ \Rightarrow \frac 12+(n-1)\times (-\frac 16)=-\frac 56 \\ \Rightarrow 6\times [\frac 12+(n-1)\times (-\frac 16)]=6 \times[-\frac 56] \\ \Rightarrow 3-(n-1)=-5\\ \Rightarrow n-1=3+5 \\ \Rightarrow n=8+1 \\ \Rightarrow n=9.$

So, the total number of terms of the series is : $\,9\,$ and the middle term is $\,\left(\frac{9+1}{2}\right)\,=5\,$ th term .

So, $\,t_{5}=\frac 12+(5-1)\times (-\frac 16) \\ \Rightarrow t_{5}=\frac 12-4 \times \frac 16 \\ \Rightarrow t_5=\frac 12-\frac 23 \\ \Rightarrow  t_5=-\frac 16\,\,\text{(ans.)}$ 

The sum of the series : 

$\,S=\frac n2(a+l) \\ \Rightarrow S=\frac{9}{2}[\frac 12+(-\frac 56)] \\ \Rightarrow S=\frac{9}{2} \times (\frac 12-\frac 56) \\ \Rightarrow S=\frac 92 \times \frac{3-5}{6} \\ \Rightarrow S= \frac 92 \times (-\frac 26) \\ \Rightarrow S=-\frac 32\,\text{(ans.)}$

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

$\,6.\,$ Find the middle term (or terms) and the sum of the each of the following series :

$\,(iii)\,~~~2+2.4+2.8+\cdots +10.4$

Sol. Here, the first term $\,(a)=2,\,$ the common difference $\,(d)=2.4-2=0.4\,$ and the last term $(l)=10.4.$

Let $\,t_n=10.4\\ \Rightarrow a+(n-1)d=10.4\\ \Rightarrow 2+(n-1)\times (0.4)=10.4 \\ \Rightarrow 0.4 \times (n-1)=10.4-2 \\ \Rightarrow n-1=\frac{8.4}{0.4} \\ \Rightarrow n=21+1=22.$

So, the total number of terms of the series is : $\,22\,$ and the middle terms are $\,\left(\frac{22}{2}\right)\,=11\,$ th term and $\,\left(\frac{22}{2}+1\right)=12\,$th term.

So, $\,t_{11}=2+(11-1)\times 0.4 \\ \Rightarrow t_{11}=2+10 \times 0.4 \\ \Rightarrow t_{11}=2+4 \\ \Rightarrow  t_{11}=6\,\,\text{(ans.)}$ 

$\,t_{12}=2+(12-1)\times 0.4 \\ \Rightarrow t_{11}=2+11 \times 0.4 \\ \Rightarrow t_{11}=2+4.4 \\ \Rightarrow  t_{11}=6.4\,\,\text{(ans.)}$ 

The sum of the series : 

$\,S=\frac n2(a+l) \\ \Rightarrow S=\frac{22}{2}[2+10.4] \\ \Rightarrow S=11 \times 12.4 \\ \Rightarrow S= 136.4\,\, \,\text{(ans.)}$

$\,7.\,$ The twelfth term of an A.P. is $\,(-13)\,$ and the sum of its first four terms is $\,24\,$ ; find the sum of its first $\,10\,$ terms of the A.P.

Sol. Here, let the first term  be denoted by $\,a,\,$ the common difference $\,d\,$ .

$\,t_{12}=a+(12-1)d \\ \Rightarrow a+11d=-13 \rightarrow(1)$

Again, $\,S_4=\frac 42[2a+(4-1)d] \\ \Rightarrow 24=2 [2a+3d]\\ \Rightarrow 2a+3d=\frac{24}{2} \\ \Rightarrow 2a+3d=12 \rightarrow(2)$

From $\,(1),\,(2)\,$ we get, 

$\,2a+3d-2(a+11d)=12-2 \times (-13) \\ \Rightarrow 3d-22d=12+26 \\ \Rightarrow -19d=38 \\ \Rightarrow d=-\frac{38}{19}=-2$

Putting the value of $\,d\,$, in $\,(1),\,$ we get,

$\,a+11 \times (-2)=-13 \\ \Rightarrow a=-13+22  \\ \Rightarrow a=9$

Hence, the sum of its first $\,10\,$ terms  denoted by $\,S_{10}\,$ and is given by :

$\,S_{10}=\frac{10}{2}[2 \times 9+(10-1)\times (-2)]\\~~~~~~=5\times [18-18]\\~~~~~~=0.$ 

$\,8.\,$  The $\,5\,$th and $\,11\,$ th terms of an A.P. are $\,41\,$ and $\,20\,$ respectively. What is its first term? Find the sum of its first eleven terms.

Sol. Here, let the first term  be denoted by $\,a,\,$ the common difference $\,d\,$ .

$\,t_{5}=a+(5-1)d \\ \Rightarrow a+4d=41 \rightarrow(1)$

Again, $\,t_{11}=a+(11-1)d \\ \Rightarrow a+10d=20 \rightarrow(2)$

From $\,(1),\,(2)\,$ we get, 

$\,a+10d-(a+4d)=20-41\\ \Rightarrow 10d-4d=-21 \\ \Rightarrow 6d=-21 \\ \Rightarrow d=-\frac{21}{6}=-\frac 72$

Putting the value of $\,d\,$, in $\,(1),\,$ we get,

$\,a+4 \times (-\frac 72)=41 \\ \Rightarrow a-14=41  \\ \Rightarrow a=41+14 \\ \Rightarrow a= 55.$

So, $\,S_{11}=\frac{11}{2}[2a+(11-1)d] \\ \Rightarrow S_{11}=\frac{11}{2} [2\times 55+10\times (-\frac 72)]\\ \Rightarrow S_{11}=\frac{11}{2}[110-35] \\ \Rightarrow S_{11}=\frac{11}{2} \times 75\\~~~~~~~~~~=\frac{825}{2}\\~~~~~~~~~~=412\frac 12\,\,\text{(ans.)}$

$\,9.\,$ The sum to $\,n\,$ terms of an A.P. is $\,n^2\,$. Find the common difference.

Sol. Here, $\,S_n=n^2; \\ S_1=a_1=1^2=1;\\ S_2=a_1+a_2=2^2=4 \\ \Rightarrow 1+a_2=4 \\ \Rightarrow a_2=4-1=3; \\ \therefore \,\, d=a_2-a_1=3-1=2\,\text{(ans.)} $

Note : Here, $\,a_1\,$ and $\,a_2\,$ denote  the first and second term of the A.P.  respectively and $\,d\,$ denotes the common difference.

$\,10.\,$ Show that the sum of $\,n\,$ terms of the series $\,\,\{4+12+20+28+\cdots\}\,\,$ is the square of an even number.

Sol. Here, the first term of the series $\,(a)=4,\,\,$ common difference $\,(d)=12-4=8.$

$\,\therefore \,\,$ the sum of $\,n\,$ terms of the series 

$=\frac n2[2a+(n-1)d]\\=\frac n2[2\times 4+(n-1)\times 8]\\=\frac n2[8+8n-8]\\=\frac n2 \times 8n\\=4n^2\\=(2n)^2 \rightarrow(1)$

Hence, from $\,(1),\,$ we can conclude that the sum of $\,n\,$ terms of the given series is the square of an even number.

$\,11.\,$  Prove that, when $\,1\,$ is added to the sum of $\,n\,$ terms of the series $\,\{8+16+24+\cdots \},\,$ the result will be a perfect square.

Sol.  Here, the first term of the series $\,(a)=8,\,\,$ common difference $\,(d)=16-8=8.$

So, $\,S_n=\frac n2[2a+(n-1)\times d]\\~~~~~=\frac n2[2\times 8+(n-1)\times 8]\\~~~~~=\frac n2(16+8n-8)\\~~~~~=\frac n2(8n+8)\\~~~~~=n(4n+4)\\~~~~~=4n^2+4\rightarrow(1)$

Hence, from $\,(1),\,$ we can say if $\,1\,$ is added to the sum of $\,n\,$ terms of the series , it becomes 

$\,4n^2+4n+1\\=(2n)^2+2\times 2n\times 1+1^2\\=(2n+1)^2 \rightarrow(2)$

Hence, from $\,(2),\,$ we can conclude that when $\,1\,$ is added to the sum of $\,n\,$ terms of the series $\,\{8+16+24+\cdots \},\,$ the result will be a perfect square.

$\,12.\,$ Find the sum of the series $\,\,1+3+4+8+7+13+10+18+\cdots \,$ to $\,23\,$ terms.

Sol. $\,\,S=1+3+4+8+7+13+10+18+\cdots \text{to 23 terms}\\=(1+4+7+10+\cdots\text{to 12 terms})\\+(3+8+13+18+\cdots\text{to 11 terms})\\=S_1+S_2 \rightarrow(1)$ 

Hence, $\,S_1=(1+4+7+10+\cdots\text{to 12 terms})\\~~~~~=\frac{12}{2}[2\times 1+(12-1)\times 3]\\~~~~~=6[2+33]\\~~~~~=6 \times 35\\~~~~~=210\rightarrow(2)$

Here, for $\,S_1,\,\,$ the first term $\,(a)=1,\,$ common difference $\,(d)=4-1=3.$

Again, $\,S_2=(3+8+13+18+\cdots\text{to 11 terms})\\~~~~~=\frac{11}{2}[2\times 3+(11-1)\times 5]\\~~~~~=\frac{11}{2}[6+50]\\~~~~~=11\times 28\\~~~~~=308\rightarrow(3)$

Hence, from $\,(1),(2),(3)\,\,$ we can say,

$\,S_1=210+308=518\,\,\text{(ans.)}$