Differentiation (Part-38) | S N De

 $\,11.\,$ The $\,n\,$-th term of an A.P. is $\,7n-5\,; \,$ find the sum of first $\,20\,$ terms of the series.

Sol. The $\,n\,$-th term of an A.P. is $\,\,t_n=7n-5.$

Now, $\,t_1=7 \times 1-5=7-5=2,\\ t_2=7 \times 2-5=14-5=9,\\ d=t_2-t_1=9-2=7.$

So, $\,S_n=\frac{n}{2}[2a+(n-1)d]\\ \therefore  S_{20}=\frac{20}{2}[2 \times 2+(20-1)\cdot7]\\~~~~~~~~~=10(4+133)\\~~~~~~~~~=1370\,\,\text{(ans.)}$

$\,12.\,$ The middle term of an arithmetic progression having $\,(2n+1)\,$ terms is $\,m;\,$ Show that the sum of its $\,(2n+1)\,$ terms is $\,(2n+1)m.$

Sol. Since the total number of terms of the given arithmetic progression is odd, there will be one middle term and that is : $\,\frac{2n+1+1}{2}=(n+1)$-th term and $\,t_{n+1}=m.$

Now, $\,\,a+(n+1-1)d=m \\ \Rightarrow a+nd=m \rightarrow(1)$

So, $\,\,S_{2n+1}=\frac{2n+1}{2}[2a+(2n+1-1)d]\\~~~~~~~~~~~=\frac{2n+1}{2}\times 2[a+nd]\\~~~~~~~~~~~=(2n+1)(a+nd)\\~~~~~~~~~~~=(2n+1)m\,\,\,[\text{By (1)}] \,\,\text{(proved)}$

$\,13(i).\,$ The sum of three integers in A.P. is $\,15\,$ and their product is $\,80;$ Find the integers.

Sol. Let the three integers be $\,\,(a-d), a , (a+d)\,\,$ so that 

$(a-d)+a+(a+d)=15 \\ \Rightarrow  3a=15 \\ \Rightarrow  a=\frac{15}{3}=5.$ 

Again, $\,(a-d).a.(a+d)=80 \\ \Rightarrow (a-d)(a+d)=\frac{80}{a} \\ \Rightarrow  a^2-d^2=\frac{80}{5} =16 \\ \Rightarrow  5^2-16=d^2 \\ \Rightarrow  d= \pm \sqrt{25-16} \\ \Rightarrow  d=\pm \sqrt{9}=\pm 3.$

Now, for $\,a=5,\,d=3\,\,$ the three numbers are 

$\,a-d=5-3=2,\\ a=5,\\a+d=5+3=8.$

Again, for $\,a=5,\,d=-3\,\,$ the three numbers are 

$\,a-d=5-(-3)=8,\\ a=5,\\a+d=5+(-3)=2.$

$\,13(ii)\,$ The sum of four integers in A.P. is $\,20\,$ and the sum of their squares is $\,120\, ;$ find them.

Sol. Let the four integers be $\,(a-3d),(a-d),(a+d),(a+3d)\,\,$ which are in A.P. with  common difference $\,2d.$

Now, $\,(a-3d)+(a-d)+(a+d)\\~~~~~~~~~~~~~~~~~+(a+3d)=20 \\ \Rightarrow 4a=20 \\ \Rightarrow a=\frac{20}{4}=5.$

Again, $\,(a-3d)^2+(a-d)^2\\~~~~~~~~+(a+d)^2+(a+3d)^2=120 \\ \Rightarrow [(a+3d)^2+(a-3d)^2]\\~~~~~~~+[(a+d)^2+(a-d)^2]=120 \\ \Rightarrow 2[a^2+(3d)^2]+2[a^2+d^2]=120 \\ \Rightarrow a^2+9d^2+a^2+d^2=\frac{120}{2} \\ \Rightarrow 2a^2+10d^2=60 \\ \Rightarrow 2\times 5^2+10d^2=60 \\ \Rightarrow 50 +10d^2=60 \\ \Rightarrow d^2=\frac{60-50}{10}=1 \\ \Rightarrow d=\pm 1. $

Now, we calculate :

$\, a-3d=5 -3 \cdot (\pm 1)=5-3=2, \\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ 5+3=8, \\ a-d=5-(\pm 1)=5-1=4,\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ 5+1=6,\\ a+d=5+ (\pm 1)=5+1=6,\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ 5-1=4, \\ a+3d=5+3(\pm 1)=5+3=8,\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ 5-3=2.$

Hence, the numbers are : $\,2,4,6,8\,\,$ or $\,\,8,6,4,2.$ 

$\,14.\,$ Divide $\,21\,$ into three parts in A.P. such that the product of first and second parts is $\,21.$

Sol. Let $\,21\,$ be divided into $\,(a-d),a,(a+d)\,\,$ so that 

$\,\,(a-d)+a+(a+d)=21 \\ \Rightarrow 3a=21 \\ \Rightarrow a=\frac{21}{3}=7.$

Now, according to the problem,

$\,(a-d)\times a=21 \\ \Rightarrow a-d=\frac{21}{a} \\ \Rightarrow 7-d=\frac{21}{7} \\ \Rightarrow 7-d=3 \\ \Rightarrow 7-3=d \\ \Rightarrow d=4.$

So, $\,21\,$ can be divided into $\,a-d=7-4=3, \\ a=7,\,a+d=7+4=11.$

$\,15.\,$ The ratio of the $\,11\,$-th and $\,14\,$-th terms of an A.P. is $\,7 : 9\,;\,$ find the ratio of the $\,10\,$-th terms to $\,3\,$-rd term of the series.

Sol. Suppose that $\,a\,$ denotes first term of A.P. and $\,d\,$ denotes the common difference of the A.P.

Now, $\,\,t_{11}=a+(11-1)d=a+10d, \\ t_{14}=a+(14-1)d=a+13d. \\ \therefore \frac{t_{11}}{t_{14}}=\frac 79 \\ \Rightarrow \frac{a+10d}{a+13d}=\frac 79 \\ \Rightarrow  9(a+10d)=7(a+13d) \\ \Rightarrow  9a+90d=7a+91d \\ \Rightarrow  9a-7a=91d-90d \\ \Rightarrow  2a=d \\ \Rightarrow  a=\frac d2.$

Now, $\,\,t_{10}=a+(10-1)d=\frac d2+9d=\frac{19d}{2}, \\ t_3=a+(3-1)d=a+2d=\frac d2+2d=\frac{5d}{2}. $

Hence, $\,t_{10} : t_3=\frac{19d}{2}: \frac{5d}{2}=19 : 5\,\,\text{(ans.)}$

$\,16.\,$ Insert five arithmetic means between $\,\,(-19)\,\,$ and $\,23.$

Sol.  If we insert five arithmetic means between $\,\,(-19)\,\,$ and $\,23,\,$ then the total numbers of terms will be $\,5+2=7\,\,$ which will be in A.P. 

Clearly, $\,t_1=a=-19,\\ t_7=23 \\ \Rightarrow a+(7-1)d=23 \\ \Rightarrow -19+6d=23 \\ \Rightarrow 6d=23+19 \\ \Rightarrow d=\frac{42}{6}=7.$ 

Hence, five arithmetic means between $\,\,(-19)\,\,$ and $\,23\,$ are 

$\,-19+7=-12,\\ -12+7=-5, \\ -5+7=2,\\ 2+7=9,\\ 9+7=16.$

$\,17.\,$ There are $\,n\,$ arithmetic means between $\,4\,$ and $\,31\,$ such that the second mean : last mean $=5 : 14\,;\,$ find $\,n.$

Sol. If there are $\,n\,$ arithmetic means between $\,4\,$ and $\,31\,$ , then there are total $\,(n+2)\,$ terms which are in A.P. and $\,\,a=t_1=4,\,t_{n+2}=31=l.$

Now,  the second mean : last mean $=5 : 14 ,$ means 

$\,t_3 : l=5: 14 \\ \Rightarrow 4+(3-1)d : l-d=5 : 14 \\ \Rightarrow \frac{4+2d}{31-d}=\frac{5}{14} \\ \Rightarrow 14(4+2d)=5(31-d) \\ \Rightarrow 56+28d=155-5d \\ \Rightarrow 28d+5d=15-56 \\ \Rightarrow 33d=99 \\ \Rightarrow d=\frac{99}{33}=3.$

Now, $\,t_{n+2}=31 \\ \Rightarrow 4+(n+2-1)d=31  \\ \Rightarrow 4+(n+1)\times 3= 31 \\ \Rightarrow 3(n+1)=31-4 \\ \Rightarrow n+1=\frac{27}{3} \\ \Rightarrow n=9-1=8.$

$\,18.\,$ If $\,a,b,c\,$ are in A.P., then show that, 

$\,(i)\,\,(a+2b-c)(2b+c-a)(c+a-b)=4abc$

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

Now, $\,\, (a+2b-c)(2b+c-a)(c+a-b)\\=(a+a+c-c)(a+c+c-a)(2b-b)\,\,[\text{By (1)}]\\=(2a)(2c)(b)\\=4abc\,\,\text{(proved)}$

$\,18.\,$ If $\,a,b,c\,$ are in A.P., then show that, 

$\,(ii)\,\,a^2(b+c),b^2(c+a), c^2(a+b)\,\,$ are in A.P. $\,\,(ab+bc+ca \neq 0)$

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

Now, if we are to show that $\,\,a^2(b+c),b^2(c+a), c^2(a+b)\,\,$ are in A.P., then we have to show that $\,\,a^2(b+c)+c^2(a+b)=2b^2(c+a).$

Now, $\,\,a^2(b+c)+c^2(a+b)\\=a^2b+a^2c+c^2a+c^2b\\=a^2b+c^2b+ac(a+c)\\=a^2b+c^2b+ac (2b)\,\,\,[\text{By (1)}]\\=b(a^2+c^2+2ac)\\=b(a+c)^2\\=b(a+c)(a+c)\\=b.2b(a+c)\\=2b^2(a+c)\rightarrow(2)$

Hence, from $\,(2),\,$ we can conclude that $\,\,a^2(b+c),b^2(c+a), c^2(a+b)\,\,$ are in A.P.

$\,18.\,$ If $\,a,b,c\,$ are in A.P., then show that, 

$\,(iii)\,\,\frac{1}{bc},\frac{1}{ca},\frac{1}{ab}\,\,$ are in A.P. 

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

Now, if we are to show that $\,\,\frac{1}{bc},\frac{1}{ca},\frac{1}{ab}\,\,$ are in A.P., then we have to show that $\,\,\frac{1}{bc}+\frac{1}{ab}=\frac{2}{ca}.$

Now, $\,\,\frac{1}{bc}+\frac{1}{ab}\\=\frac{a+c}{abc}\\=\frac{2b}{abc}\,\,\,[\text{By (1)}]\\=\frac{2}{ac}\rightarrow(2)$

Hence, from $\,(2),\,$ we can conclude that $\,\frac{1}{bc},\frac{1}{ca},\frac{1}{ab}\,\,$ are in A.P.