Differentiation (Part-38) | S N De

 $\,8.\,$ The sums of first $\,n\,$ terms of two A.P.'s are in the ratio $\,\,(3n+5): (5n-9).\,$ Show that their $\,4\,$-th terms are equal.

Sol.  By question, $\,\,\frac{\frac n2[2a_1+(n-1)d_1]}{\frac n2[2a_2+(n-1)d_2]}=\frac{3n+5}{5n-9}\,[*]\\ \Rightarrow \frac{2a_1+(n-1)d_1}{2a_2+(n-1)d_2}=\frac{3n+5}{5n-9}\rightarrow(1)$

Now, putting $\,\,n=7,\,\,$we get from $\,(1),\,$

$\,\frac{2a_1+(7-1)a_1}{2a_2+(7-1)d_2}=\frac{3\times 7+5}{5\times 7-9} \\ \Rightarrow \frac{a_1+3d_1}{a_2+3d_2}=\frac{26}{26} \\ \Rightarrow\frac{a_1+(4-1)d_1}{a_2+(4-1)d_2}=1 \\ \Rightarrow a_1+(4-1)d_1=a_2+(4-1)d_2\rightarrow(2)$

From $\,(2),\,$we can say that their $\,4\,$-th terms are equal.

Note[*] : $\,\,a_1,a_2\,$ indicate the first terms of two A.P.s, $\,d_1,d_2\,$ denote the common differences of two different A.P.s.

$\,9.\,$ Insert a number of arithmetic means between between $\,4\,$ and $\,34\,$ such that the sum of the resulting A.P. is $\,133.$

Sol. Let $\,n\,$ number of arithmetic means between between $\,4\,$ and $\,34\,$ are inserted so that total number of terms of A.P. are $\,n+2.$

$\,\therefore \frac{n+2}{2}(4+34)=133\\ \Rightarrow \frac{n+2}{2}\times 38=133 \\ \Rightarrow 19(n+2)=133 \\ \Rightarrow n+2=\frac{133}{19}=7 \\ \Rightarrow n=5.$

So, total number of terms of the given A.P. are $\,n+2=5+2=7.$

Now, $\,4+(7-1)d=34 \\ \Rightarrow d=5.$

Hence, the arithmatic means are given by : 

$\,\,4+5=9, \\ \,4+2\times 5=14, \\ \, 4+3 \times 5=19, \\ \, 4+4 \times 5=24,\\ \,4+ 5 \times 5=19.$

$\,10.\,$ If $\,a,b,c\,$ are in A.P. then show that , $\,a\left(\frac 1b+\frac 1c\right),\,b\left(\frac 1c+\frac 1a\right),c\left(\frac 1a+\frac 1b\right)\,\,$ are also in A.P.

Sol. Since $\,a,b,c\,$ are in A.P., $\,2b=a+c\rightarrow(1).$

Now, $\,a\left(\frac 1b+\frac 1c\right)+c\left(\frac 1a+\frac 1b\right)\\=a.\left(\frac{c+b}{bc}\right)+c.\left(\frac{b+a}{ab}\right)\\=\frac{a^2c+a^2b+bc^2+ac^2}{abc}\\=\frac{b(a^2+c^2)+ac(a+c)}{abc}\\=\frac{b[(a+c)^2-2ac]+ac(a+c)}{abc}\\=\frac{b[(2b)^2-2ac]+ac(2b)}{abc}\,\,\text{[By (1)]} \\=\frac{b(4b^2-2ac)+2abc}{abc}\\=\frac{b.4b^2-2abc+2abc}{abc}\\=\frac{2b.2b}{ac}\\=\frac{2b(a+c)}{ac}\,\,\text{[By (1)]}\\=2b.\frac{a+c}{ac}\\=2b\left(\frac 1c+\frac 1a\right)\,\,\,\text{(showed)}$

$\,11.\,$ Three positive numbers $\,a,b,c\,$ are in A.P. Prove that, $\,\frac{1}{\sqrt b+\sqrt c},\,\frac{1}{\sqrt c+\sqrt a},\,\frac{1}{\sqrt a+\sqrt b}\,\,$ are also in A.P.

Sol. Since three positive numbers $\,a,b,c\,$ are in A.P., $\,a-b=b-c\rightarrow(1)$

Now, $\,\,\frac{}{\sqrt b+\sqrt c}+\frac{1}{\sqrt a+\sqrt b}\\=\frac{\sqrt b-\sqrt c}{b-c}+\frac{\sqrt a-\sqrt b}{a-b}\\=\frac{\sqrt b-\sqrt c+\sqrt a-\sqrt b}{b-c}\,\,\quad \text{[By (1)]}\\=\frac{-\sqrt c+\sqrt a}{b-c}\\=\frac{\sqrt c-\sqrt a}{c-b}\\=\frac{\sqrt c-\sqrt a}{c-\frac{a+c}{2}}\,\,[\because 2b=c+a\,\text{(By (1))}]\\=2.\frac{\sqrt c-\sqrt a}{2c-a-c}\\=2.\frac{\sqrt c-\sqrt a}{c-a}\\=2.\frac{(\sqrt c-\sqrt a)}{(\sqrt c+\sqrt a)(\sqrt c-\sqrt a)}\\=2.\frac{1}{\sqrt c+\sqrt a}\,\,\,\text{(proved)}$

$\,12.\,$ If $\,\,a_1,a_2,a_3,\cdots, a_n\,\,$ be in A.P. Show that , $\,\,\frac{1}{a_1a_2}+\frac{1}{a_2a_3}+\cdots+\frac{1}{a_{n-1}a_n}=\frac{n-1}{a_1a_n}$

Sol. $\,\,\because \,\,a_1,a_2,a_3,\cdots, a_n\,\,$ are in A.P., 

$\,d=a_2-a_1=a_3-a_2=\cdots=a_n-a_{n-1}.$

Now, $\,\,\frac{1}{a_1a_2}+\frac{1}{a_1a_2}+\cdots+\frac{1}{a_{n-1}a_n}\\=\frac 1d\left[\frac{a_2-a_1}{a_1a_2}+\frac{a_3-a_2}{a_2a_3}+\cdots +\frac{a_n-a_{n-1}}{a_{n-1}a_n}\right]\\=\frac 1d \left[\frac{1}{a_1}-\frac{1}{a_2}+\frac{1}{a_2}-\frac{1}{a_3}+\cdots+\frac{1}{a_{n-1}}-\frac{1}{a_n}\right]\\=\frac 1d\left(\frac{1}{a_1}-\frac{1}{a_{n}}\right)\\=\frac 1d \left(\frac{a_n-a_1}{a_1a_n}\right)\\=\frac 1d \times \frac{(n-1)d}{a_1a_n}\,\,[*]\\=\frac{n-1}{a_1a_n}\,\,\text{(showed)}$

Note[*] : $\,\,a_n=a_1+(n-1)d,\\~~~~\text{where d is common difference.}$

$\,13.\,$ If $\,(b-c)^2,\,(c-a)^2,\,(a-b)^2\,$ are in A.P. then prove that $\,\,\frac{1}{b-c},\,\frac{1}{c-a},\,\frac{1}{a-b}\,$ are also in A.P.

Sol. By question, $\,(c-a)^2-(b-c)^2=(a-b)^2-(c-a)^2 \\ \Rightarrow (c-a+b-c)(c-a-b+c)\\=(a-b+c-a)(a-b-c+a)\\ \Rightarrow -(a-b)(2c-a-b)\\=-(b-c)(2a-b-c) \\ \Rightarrow \frac{(a-b)(2c-a-b)}{(a-b)(b-c)(c-a)}=\frac{(b-c)(2a-b-c)}{(a-b)(b-c)(c-a)} \\ \Rightarrow \frac{(c-a)-(b-c)}{(b-c)(c-a)}=\frac{(a-b)-(c-a)}{(a-b)(c-a)} \\ \Rightarrow \frac{1}{b-c}-\frac{1}{c-a}=\frac{1}{c-a}-\frac{1}{a-b}\\ \Rightarrow \frac{1}{b-c}+\frac{1}{a-b}=\frac{2}{c-a} \rightarrow(1)$

Hence , from $\,(1),\,$ we can conclude that $\,\,\frac{1}{b-c},\,\frac{1}{c-a},\,\frac{1}{a-b}\,$ are  in A.P.

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

$\,(i)\,\,2 \cdot 4 +6 \cdot 8+ 10 \cdot 12+ \cdots$

Sol.  $\,t_r=[2+(r-1)\cdot 4][4+(r-1)\cdot 4]\\~~~~=(4r-2)(4r)\\~~~~=16r^2-8r$

The sum to $\,n\,$ terms is given by :

$\,\,\sum_{r=1}^n (16r^2-8r)\\=16\sum_{r=1}^n r^2-8\sum_{r=1}^n r \\=16 \times  \frac{n(n+1)(2n+1)}{6}-8 \times \frac{n(n+1)}{2}\\=\frac 83n(n+1)(2n+1)-4n(n+1)\\=4n(n+1)\left[\frac 23(2n+1)-1\right]\\=\frac{4n(n+1)(4n-1)}{3}\,\,\text{(ans.)}$

$\,14(ii)\,\, 1^2+3^2+5^2+\cdots $

Sol. $\,\,t_r=(2r-1)^2=4r^2-4r+1,\\ [\text{where}\,\,\,t_r\,\,\text{being r-th term}]$ 

So, $\,\,\sum_{r=1}^n(4r^2-4r+1)\\=4\sum_{r=1}^n r^2-4\sum_{r=1}^n r+\sum_{r=1}^n 1\\=4 \cdot \frac{n(n+1)(2n+1)}{6}-4 \cdot \frac{n(n+1)}{2}+n\\=\frac 23n(n+1)(2n+1)-2n(n+1)+n\\=2n(n+1)\left[\frac{2n+1}{3}-1\right]+n\\=2n(n+1)\frac{2n+1-3}{3}+n\\=2n(n+1)\frac{2n-2}{3}+n\\=\frac 43 n(n+1)(n-1)+n\\=\frac 43n(n^2-1)+n\\=n\left[\frac{4n^2-4}{3}+1\right]\\=\frac n3(4n^2-1)\,\,\text{(ans.)}$ 

$\,14(iii)\,\,1 \cdot 3 \cdot5 + 3 \cdot 5\cdot 7+ 5 \cdot 7 \cdot 9+ \cdots$

Sol. $\,\,t_r=(2r-1)(2r+1)(2r+3)\\~~~~~=8r^3+12r^2-2r-3 $

So, $\,\sum_{r=1}^n (8r^3+12r^2-2r-3)\\=8 \sum_{r=1}^n  r^3+12\sum_{r=1}^n  r^2-2\sum_{r=1}^n  r \\-\sum_{r=1}^n  3\\=8 \left[\frac{n(n+1)}{2}\right]^2+12 \cdot \frac{n(n+1)(2n+1)}{6}\\-2 \cdot \frac{n(n+1)}{2}-3n \\=2n^2(n+1)^2+2n(n+1)(2n+1)\\-n(n+1)-3n\\=2n(n+1)(n^2+n+2n+1)-n(n+1)\\-3n\\=2n(n+1)(n^2+3n+1)-n(n+1)-3n\\=n(n+1)(2n^2+6n+2-1)-3n\\=n(n+1)(2n^2+6n+1)-3n\,\,\text{(ans.)}$