Differentiation (Part-38) | S N De

 $\,23.\,$ If $\,S_n\,$ be the sum of $\,n\,$ consecutive terms of an A.P. show that , 

$\,(i)\,\, S_{n+3}-3S_{n+2}+3S_{n+1}-S_n=0$

Sol Let $\,d\,$ be the common difference of the A.P. 

So, $\,\,\,t_{n+3}-t_{n+2}=t_{n+2}-t_{n+1}=d\rightarrow(1)$

Now, $\,\,S_{n+3}-3S_{n+2}+3S_{n+1}-S_n \\=(S_{n+3}-S_{n+2})-2(S_{n+2}-S_{n+1})\\ ~~~~~~~+(S_{n+1}-S_n)\\=t_{n+3}-2t_{n+2}+t_{n+1}\\=(t_{n+3}-t_{n+2})-(t_{n+2}-t_{n+1})\\=d-d\,\,\text{[By (1)]}\\=0\,\,\text{(showed)}$

$\,23.\,$ If $\,S_n\,$ be the sum of $\,n\,$ consecutive terms of an A.P. show that , 

$\,(ii)\,\, S_{n+4}-4S_{n+3}+6S_{n+2}-4S_{n+1}\\ +S_n=0$

Sol. Let $\,d\,$ be the common difference of the A.P. 

So, $\,\,\,t_{n+4}-t_{n+3}=t_{n+3}-t_{n+2}\\ =t_{n+2}-t_{n+1}=d\rightarrow(1)$

Now, $\,\,S_{n+4}-4S_{n+3}+6S_{n+2}-4S_{n+1}+S_n \\=(S_{n+4}-S_{n+3})-3(S_{n+3}-S_{n+2})\\ +3(S_{n+2}-S_{n+1})-(S_{n+1}-S_{n})\\=t_{n+4}-3t_{n+3}+3t_{n+2}-t_{n+1}\\=(t_{n+4}-t_{n+3})-2(t_{n+3}-t_{n+2})\\ +(t_{n+2}-t_{n+1})\\=d-2d+d\,\,\text{[By (1)]}\\=0\,\,\text{(showed)}$

$\,24.\,$ Let $\,a_n\,$ denote the $\,n\,$-th term of an A.P. and $\,p\,$ and $\,q\,$ be two positive integers with $\,p<q.\,\,$ If $\,\,a_{p+1}+a_{p+2}+a_{p+3}\cdots +a_{q}=0\,\,$ find the sum of first $\,(p+q)\,$ terms of the A.P.

Sol.  $\,\,a_{p+1}=a_{1}+(p+1-1)d=a_1+pd, \\ a_{p+2}=a_{1}+(p+2-1)d=a_{1}+(p+1)d, \\ a_{p+3}=a_{1}+(p+3-1)d=a_{1}+(p+2)d, \\ \vdots \\ a_{q}=a_{1}+(q-1)d.$

So,  $\,\,a_{p+1}+a_{p+2}+a_{p+3}\cdots +a_{q}\\=(q-p)a_{1}+d[p+(p+1)+(p+2) \\+\cdots+(q-1)]\\=(q-p)a_{1} \\+d \times \left[\frac{q-p}{2}\{2p+(q-p-q)\times 1\}\right]\\=(q-p)a_{1} \\+d \times \left[(q-p)p+\frac 12(q-p)(q-p-1)\right]\\=\frac{q-p}{2}[2a_1+2pd+(q-p-1)d]\\=\frac{q-p}{2}[2a_1+(2p+q-p-1)d]\\=\frac{q-p}{2}[2a_1+(p+q-1)d]\rightarrow(1)$

Now,  $\,\,a_{p+1}+a_{p+2}+a_{p+3}\cdots +a_{q}=0\,\,$ gives 

$\,2a_1+(p+q-1)d=0\,\,\,\text{[By (1)]}$ 

So, $\,\,S_{p+q}=\frac{p+q}{2}[2a_1+(p+q-1)d]\\ \Rightarrow S_{p+q}=\frac{p+q}{2} \times 0=0 \,\,\text{(ans.)}$

$\,25.\,$ Find the sum of the three-digited natural numbers which leave a remainder $\,2\,$ , when divided by $\,3.$

Sol. By question, the three-digited natural numbers are $\,101,104,107,\cdots, 998\,\,$ 

Again, $\,998=101+(n-1) \times 3 \,\,[*] \\ \Rightarrow (998-101)=3(n-1) \\ \Rightarrow 897=3(n-1) \\ \Rightarrow n-1=\frac{897}{3} \\ \Rightarrow n=299+1=300$

So, $\,S=\frac{300}{2}[101+998] \\ \Rightarrow S=150 \times 1099=164850\,\,\text{(ans.)}$

Note[*]: $\,t_n=a+(n-1)d,\,$ where  $\,a\,$ denotes first term of A.P. and $\,d\,$ denotes the common difference of the A.P.

$\,26.\,$ Between the roots of the  quadratic equation $\,\,3px^2-10px+5q=0\,\,(p>0,\,\frac qp <1\frac 23),\,\,$ an odd number of A.M.s are inserted and their sum exceeds their number by $\,10.$ Find the number of A.M.s inserted.

Sol. Let the roots of the  quadratic equation $\,\,3px^2-10px+5q=0\,\,(p>0,\,\frac qp <1\frac 23),\,\,$  are $\,l,m$ and $\,n\,$ number of A.M.s are inserted in between them so that 

$\,l+m=-\frac{(-10p)}{3p}=\frac{10}{3}\rightarrow(1)$ 

Since, $\,n\,$ is odd, we can take $\,n=2k-1.$

By question, let $\,a_1,a_2,\cdots, a_n$ be $\,n\,$ A.M.s between $\,l,m.$

Now, $\,a_1+a_2+\cdots +a_n=n+10 \rightarrow(2)$

So, we can see that the sequence $\,\,l,a_1,a_2,\cdots, a_n,m\,$ form a sequence with  $\,(n+2)\,$ terms and these terms are in A.P.

$\,l+a_1+a_2+\cdots+a_n+m=\frac{n+2}{2} \times (l+m) \\ \Rightarrow (l+m)+(a_1+a_2+\cdots +a_n) \\ =\frac{2k-1+2}{2}\times \frac{10}{3}\,\,\text{[By (1)]} \\ \Rightarrow \frac{10}{3}+(n+10)=\frac{2k+1}{2} \times \frac{10}{3} \,\,\text{[By (2)]}  \\ \Rightarrow \frac{10}{3}+2k-1+10=\frac 53 \times (2k+1) \\ \Rightarrow \frac{10}{3}+2k+9=\frac{10k}{3}+\frac 53 \\ \Rightarrow \frac{37}{3}-\frac 53=\frac{10k}{3}-2k \\ \Rightarrow \frac{10k-6k}{3}=\frac{37-5}{3} \\ \Rightarrow 4k=32 \\ \Rightarrow k=\frac{32}{4}=8.$

So, $\,n=2k-1=2\times 8-1=15$

Hence, the number of A.M.s inserted  is $\,\,15.$

$\,27.\,$ Let $\,a_n\,$ be the $\,n\,$-th term of an A.P. and $\,\,a_3+a_5+a_{8}+a_{14}+a_{17}+a_{19}=198.\,\,$ Find the sum of first $\,21\,$ terms of the A.P.

Sol. Let $\,l\,$ be the first term of the A.P. with common difference $\,\,d,\,$ so that $\,a_3=l+(3-1)d=l+2d, \\ a_5=l+(5-1)d=l+4d,\\a_8=l+(8-1)d=l+7d, \\ a_{14}=l+(14-1)d=l+13d \\ a_{17}=l+(17-1)d=l+16d \\ a_{19}=l+(19-1)d=l+18d.$

Now, $\,\,a_3+a_5+a_{8}+a_{14}+a_{17}+a_{19}=198\\ \Rightarrow (l+2d)+(l+4d)+(l+7d) \\ +(l+13d)+(l+16d)+(l+18d)=198 \\ \Rightarrow 6l+60d=198 \\ \Rightarrow l+10d =\frac{198}{6}=33 \rightarrow(1)$

So, $\,S_{21}=\frac{21}{2}[l+l+(21-1)d]\\ \Rightarrow S_{21}=\frac{21}{2} [2l+20d]\\~~~~~~~~~~~= 21 \times (l+10d)\\~~~~~~~~~~~= 21 \times 33\,\,[\text{By (1)}]\\~~~~~~~~~~~= 693\,\,\,\text{(ans.)}$

$\,28.\,$ Find the sum upto the $\,n\,$-th term of the series $\,(1+3)+(5+7+9+11)\\ +(13+15+17+19+21+23)+\cdots$

Sol. Total number of terms $=2+4+6+\cdots \text{ to n terms}\\=2(1+2+3+\cdots \text{to n terms})\\=2\times \frac{n(n+1)}{2}\\=n(n+1)$

So, if we open up the parentheses then the series turn out to be 

$\,\,1+3+5+7+9+11 \\ +\cdots \text{to n(n+1) terms}\\=\frac{n(n+1)}{2}[2 \times 1+\{n(n+1)-1\} \times 2]\\=n(n+1)[1+n(n+1)-1]\\=n^2(n+1)^2\,\,\text{(ans.)}$

$\,29.\,$ If $\,S_1, S_2,S_3\,$ are the sums of $\,n\,$ natural numbers, their squares , their cubes respectively show that $\,9S_2^2=S_3(1+8S_1).$

Sol. $\,\,S_1=1+2+3+\cdots +n=\frac{n(n+1)}{2}, \\ S_2=1^2+2^2+3^2+\cdots +n^2=\frac{n(n+1)(2n+1)}{6},\\ S_3=1^3+2^3+3^3+\cdots+n^3=\left(\frac{n(n+1)}{2}\right)^2$

Now, $\,S_3[1+8S_1]\\=\frac{n^2(n+1)^2}{4} \times \left[1+8 \times \frac{n(n+1)}{2}\right]\\=\frac{n^2(n+1)^2}{4}\times [1+4n^2+4n]\\=\frac 14n^2(n+1)^2 \times (2n+1)^2\rightarrow(1)$

Again, $\,\,9S_2^2=9 \times \frac{n^2(n+1)^2(2n+1)^2}{36}\\~~~~~~~~=\frac{1}{4}n^2(n+1)^2(2n+1)^2\rightarrow(2)$

Hence, from $\,(1)\,$ and $\,(2)\,$ the result follows.

$\,30.\,$ If $\,a_1=2\,\,$ and $\,a_n-a_{n-1}=2n\,\,(n \geq 2),\,\,$ find the value of $\,\,a_1+a_2+a_3+\cdots+a_{20}.$

Sol. We have, $\,a_n-a_{n-1}=2n,\,\,a_1=2 \\ \therefore a_2-a_1=2\times 2 \\ \Rightarrow a_2-2=4 \Rightarrow a_2=6 \\ \therefore a_3-a_2=2 \times 3 \\ \Rightarrow a_3-6=6 \Rightarrow a_{3}=12\\ \text{Similarly,}\,\,a_4=20.$

Now, $\,\,a_1+a_2+a_3+\cdots+a_{20}\\=2+6+12+20+\cdots+a_{20}\\=(1^2+1)+(2^2+2)+(3^2+3) \\ +\cdots+(20^2+20)\\=(1^2+2^2+3^2+\cdots+20^2) \\ +(1+2+3+\cdots+20)\\=\frac{20 \cdot(20+1)\cdot ( 2\times 20+1)}{6}+\frac{20 \cdot (20+1)}{2}\\=\frac{20 \cdot 21 \cdot 41}{6}+10 \cdot 21\\=3080\,\,\text{(ans.)}$

$\,31.\,$ Find the middle term of the series $\,\,1+5+9+\cdots +101.$

Sol. Let the number of terms of the given series is : $\,n. $

So, $\,\,t_n=a+(n-1)d \\ \Rightarrow 101=1+(n-1)\times 4 \\ \Rightarrow n-1=\frac{101-1}{4}\\ \Rightarrow n=25+1=26.$

Hence , the middle term $\,t_{13},\,\,t_{14},\,\,\,$ where $\,t_n\,$ denotes $\,n\,$-th term.

So, $\,t_{13}=1+(13-1)\cdot 4=1+48=49, \\ t_{14}=1+(14-1)\cdot 4=53.$

$\,32.\,$ If $\,\frac{b^2+c^2-a^2}{2bc},\,\,\frac{c^2+a^2-b^2}{2ca},\,\,\frac{a^2+b^2-c^2}{2ab}\,\,$ are in A.P. and $\,\,a+b+c=0\,\,$ then prove that $\,\,a(b+c-a),\,b(c+a-b),\,c(a+b-c)\,\,$ are in A.P.

Sol. $\quad  \because \,\frac{b^2+c^2-a^2}{2bc},\,\,\frac{c^2+a^2-b^2}{2ca},\,\,\frac{a^2+b^2-c^2}{2ab}\,\,$ are in A.P., 

$\,\, \therefore \frac{b^2+c^2-a^2}{2bc}+1,\,\,\frac{c^2+a^2-b^2}{2ca}+1,\,\,\frac{a^2+b^2-c^2}{2ab}+1\,\,$ are in A.P. 

$\,\, \therefore \frac{(b+c)^2-a^2}{2bc},\,\,\frac{(c+a)^2-b^2}{2ca},\,\,\frac{(a+b)^2-c^2}{2ab}\,\,$ are in A.P. 

$\,\, \therefore \frac{(b+c+a)(b+c-a)}{2bc},\,\,\frac{(c+a+b)(c+a-b)}{2ca},\,\,\frac{(a+b+c)(a+b-c)}{2ab}\,\,$ are in A.P. 

$\,\, \therefore \frac{(b+c-a)}{2bc},\,\,\frac{(c+a-b)}{2ca},\,\,\frac{(a+b-c)}{2ab}\,\,$ are in A.P. 

$\,\, \therefore \frac{a(b+c-a)}{2abc},\,\,\frac{b(c+a-b)}{2abc},\,\,\frac{c(a+b-c)}{2abc}\,\,$ are in A.P. 

$\,\, \therefore a(b+c-a),\,\,b(c+a-b),\,\,c(a+b-c)\,\,$ are in A.P. (proved)