Differentiation (Part-38) | S N De

 

$\,19.\,$ If the $\,p\,$-th , $\,q\,$-th and $\,r\,$-th terms of an A.P. be $\,P, Q, R\,$ respectively, then show that , $\,p(Q-R)+q(R-P)+r(P-Q)=0.$

Sol. Let $\,a\,$ be the first term and $\,d\,$ be the common difference of the given A.P.

So, $\,t_p=P=a+(p-1)d,\\t_q=Q=a+(q-1)d,\\ t_r=R=a+(r-1)d.$

Now, $\,Q-R\\=[a+(q-1)d]-[a+(r-1)d]\\=(q-r)d.$

Similarly, $\,R-P=(r-p)d,\\ P-Q=(p-q)d.$

Now, $\,p(Q-R)+q(R-P)+r(P-Q)\\=p\times (q-r)d+q \times (r-p)d+r \times (p-q)d\\=d \times [p(q-r)+q(r-p)+r(p-q)]\\=d \times [pq-pr+qr-pq+rp-qr]\\=d \times 0\\=0\,\,\text{(proved)}$

$\,20(i)\,$ If $\,\frac{b+c}{a},\frac{c+a}{b},\frac{a+b}{c}\,\,$ are in A.P. show that $\,\frac 1a,\frac 1b,\frac 1c\,\,$ are also in A.P. $\,\,(a+b+c \neq 0)$

Sol. $\,\frac{b+c}{a},\frac{c+a}{b},\frac{a+b}{c}\,\,$ are in A.P. (Given)

$\Rightarrow \,\left(\frac{b+c}{a}+1\right),\left(\frac{c+a}{b}+1\right),\left(\frac{a+b}{c}+1\right)\,\,$ are in A.P.

$\Rightarrow \,\left(\frac{b+c+a}{a}\right),\left(\frac{c+a+b}{b}\right),\left(\frac{a+b+c}{c}\right)\,\,$ are in A.P.

$\Rightarrow \,\left[\frac{b+c+a}{a(a+b+c)}\right],\left[\frac{c+a+b}{b(a+b+c)}\right],\left[\frac{a+b+c}{c(a+b+c)}\right]\,\,$ are in A.P.

$\Rightarrow \frac{1}{a},\frac{1}{b},\frac{1}{c}\,\,$ are in A.P. $[\,\because\,a+b+c \neq 0]$

$\,20(ii)\,$ Let $\,a(a \neq 0)\,\,$ is a fixed real number and $\,\frac{a-x}{px}=\frac{a-y}{qy}=\frac{a-z}{rz}.\,\,$ If $\,p,q,r\,$ are in A.P. , show that $\,\frac 1x,\frac 1y,\frac 1z\,$ are in A.P.

Sol.  Let $\,\frac{a-x}{px}=\frac{a-y}{qy}=\frac{a-z}{rz}= m(\neq 0) \\ \therefore \frac{a-x}{px}=m \Rightarrow p=\frac{a-x}{mx}, \\~~~~ \frac{a-y}{qy}=m \Rightarrow q=\frac{a-y}{my}, \\ ~~~~\frac{a-z}{rz}=m \Rightarrow r=\frac{a-z}{mz}.$

If $\,p,q,r\,$ are in A.P. , $\,\,p+r=2q \\ \Rightarrow \frac{a-x}{mx}+\frac{a-z}{mz}=\frac{2(a-y)}{my} \\ \Rightarrow \frac{a-x}{x}+\frac{a-z}{z}=\frac{2(a-y)}{y} \\ \Rightarrow \frac ax-1+ \frac az-1=2\left(\frac ay-1\right) \\ \Rightarrow \frac ax+\frac az-2=\frac{2a}{y}-2 \\ \Rightarrow \frac ax+\frac az=\frac{2a}{y} \\ \Rightarrow \frac ix+\frac 1z=\frac{2}{y} \rightarrow(1)$

Hence, from $\,(1),\,$ we can conclude that $\,\frac 1x,\frac 1y,\frac 1z\,$ are in A.P.

$\,20(iii)\,$ If $\,a+c=2b\,\,$ and $\,ab+cd+ad=3bc,\,\,$ prove that the four numbers $\,a,b,c,d\,$ are in A.P.  $\,\,(b \neq 0)$

Sol.  $\,a+c=2b\,\,\text{(Given)} \\ \Rightarrow a+c=b+b \\ \Rightarrow c-b=b-a \rightarrow(1)$

Again, $\,ab+cd+ad=3bc \\ \Rightarrow cd+ad=3bc-ab \\ \Rightarrow d(c+a)=b(3c-a) \\ \Rightarrow d \times 2b=b(3c-a) \\ \Rightarrow d=\frac{3c-a}{2} \\ \Rightarrow d-c=\frac{3c-a}{2} -c \\ \Rightarrow d-c=\frac{3c-a-2c}{2} \\ \Rightarrow d-c=\frac{c-a}{2}  \\ \Rightarrow d-c=\frac{2c-(a+c)}{2}  \\ \Rightarrow d-c=\frac{2c-2b}{2}=c-b\rightarrow(2)$

Hence, from $\,(1),\,(2)\,\,$ it follows that 

$\,b-a=c-b=d-c\,\,$ and so the four numbers $\,a,b,c,d\,$ are in A.P. 

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

$\,21.\,$ Find the least value of $\,n\,$ for which the sum of the series $\,\, 20+28+36+\cdots\text{to n terms}\,\,$ is greater than $\,1000.$

Sol. Consider the given series $\,\, 20+28+36+\cdots\text{to n terms}\,\,$ where first term $(a)=20,\,$ common difference $(d)=8.$

So, $\,S_n=\frac{n}{2}[2 \times 20+(n-1)\times 8]\\~~~~~=n[20+(n-1)\times 4]\\~~~~~=n(16+4n)$

Now, from the given condition, $\, S_n >1000 \\ \Rightarrow  n(16+4n)>1000 \\ \Rightarrow 4n^2+16n >1000 \\ \Rightarrow 4(n^2+4n)> 4 \times 250 \\ \Rightarrow  n^2+4n >250 \\ \Rightarrow  n^2+4n-250 >0 \rightarrow(1)$

If we take $\,P(n)= n^2+4n-250 ,$ then we notice $\,P(13)<0\,$ but $\,P(14)>0$ and so the least value of $\,n\,$ for which the sum of the series $\,\, 20+28+36+\cdots\text{to n terms}\,\,$ is greater than $\,1000\,$ is $\,14.$

Alternatively, from $\,(1),\,$ we get , $\,n^2+4n+4>250+4 \\ \Rightarrow (n+2)^2>254 \rightarrow(2)$

Now the next square number after $\,254\,$ is $\,256=16^2\,$ so that from $\,(2),\,$ we get,

$\,\, (n+2)^2=256=16^2 \\ \Rightarrow n+2=16 \\ \Rightarrow n=16-2=14\,\text{(ans)}$

$\,22.\,$ The $\,n\,$-th term of a series in A.P. is $\,an+b.\,\,$ Find the sum of the series up to $\,\,n\,$ terms. 

Sol. Here, $\,t_n=an+b\,\,\text{(Given)}$

So, $\,\,t_1+t_2+t_3+\cdots+t_n\\=(a\cdot 1+b)+(a \cdot 2+b)+(a \cdot 3+b)\\+\cdots +(a\cdot n +b)\\=a(1+2+3+\cdots +n)+nb\\=a\cdot \frac{n(n+1)}{2}+nb\\=\frac n2[a(n+1)+2b]\,\,\text{(ans.)}$

$\,23.\,$ Find the sum of the series $\,\, 1^2-2^2+3^2-4^2+5^2-6^2+\cdots $ up to $\,2n\,$ terms.

Sol. $\,\,S_n= 1^2-2^2+3^2-4^2+5^2-6^2\\ +\cdots \text{ up to 2n terms}\\~~~~~~=(1-4)+(9-16)+(25-36) \\ +\cdots \text{ up to n terms}\\~~~~~~=-3+(-7)+(-11) \\ +\cdots \text{ up to n terms} \rightarrow(1)$

Now, from $\,(1),\,$ we see the above series is in an A.P. with first term $(a)=-3\,\,$ and common difference $\,(d)=-4.$

So, $\,S_n=\frac n2[2\times (-3)+(n-1)\times (-4)]\\~~~~~=n[-3-2(n-1)]\\~~~~~=n(-2n-1)\\~~~~~=-n(2n+1)\,\,\text{(ans.)}$ 

$\,24.\,$ The perpendicular of a right angled triangle is $\,9\,$ cm and the three sides are in A.P. Find the integral value of the length of the hypotenuse.

Sol.  Case -$1$ : Let the perpendicular of a right angled triangle  $(\,9\, \text{cm} )\,\,$ is smallest. Then, the other two sides being $\,(9+d)\,$ cm and $\,(9+2d)\,$ cm. $\,d\,$ being the common difference.

So, $\,(9+2d)^2=9^2+(9+d)^2 \\ \Rightarrow (9+2d)^2-(9+d)^2=81 \\ \Rightarrow (9+2d+9+d)(9+2d-9-d)=81 \\ \Rightarrow (18+3d)(d)=81 \\ \Rightarrow 3(6+d)d=81 \\ \Rightarrow d^2+6d=\frac{81}{3} \\ \Rightarrow d^2+6d-27=0 \\ \Rightarrow d^2+9d-3d-27=0 \\ \Rightarrow d(d+9)-3(d+9)=0 \\ \Rightarrow (d+9)(d-3)=0 \\ \Rightarrow d-3=0 [\,\,\because d+9 \neq 0] \\ \Rightarrow d=3.$

Hence, the length of the hypotenuse 

$=9+2d\\=9+2 \cdot 3\\=9+6\\=15\,\text{cm.(ans.)}$

Case-$2 :$  Let the smallest side of triangle be base . 

Suppose the three sides are $\,(9-d)\,\text{cm},9\,\text{cm},(9+d)\,\text{cm},~~\,d\,$ being the common difference.

So, $\,(9+d)^2=9^2+(9-d)^2 \\ \Rightarrow (9+d)^2-(9-d)^2=81  \\ \Rightarrow 4 \times 9 \times d=81  \\ \Rightarrow 36d=81  \\ \Rightarrow d=\frac{81}{36}=\frac 94.$

So, the length of the hypotenuse $=9+\frac 94=\frac{45}{4}\,\text{cm}\,$ which is not possible as we have to find the integral value of the hypotenuse.

$\,25.\,$ If the $\,x\,$-th term of an A.P. is $\,\frac 1y\,$ and its $\,y\,$-th term is $\,\frac 1x\,$ then show that its $\,xy\,$-th term is $\,1\,$ and the sum of its first $\,xy\,$-th term is $\,\frac 12(xy+1).$

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

Now, $\,t_x=\frac 1y \\ \Rightarrow a+(x-1)d=\frac 1y \rightarrow(1) \\ \text{and}\,\,\,\,t_y=\frac 1x \\ \Rightarrow a+(y-1)d=\frac 1x \rightarrow(2)$

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

$\,\,(x-y)d=\frac 1y-\frac 1x \\ \Rightarrow (x-y)d=\frac{x-y}{xy} \\ \Rightarrow  d=\frac{1}{xy}.$

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

$\,a+(x-1) \times \frac{1}{xy}=\frac 1y \\ \Rightarrow  a+\frac{x-1}{xy}=\frac 1y \\ \Rightarrow a+\left(\frac 1y-\frac{1}{xy}\right)=\frac 1y \\ \Rightarrow  a=\frac{1}{xy}.$

Hence, $\,t_{xy}=a+(xy-1)d \\~~~~~~ =\frac{1}{xy}+(xy-1)\frac{1}{xy}\\~~~~~~=\frac{1}{xy} (1+xy-1)\\~~~~~~=\frac{1}{xy} \times (xy)\\~~~~~~=1$

Also, $\,S_{xy}=\frac{xy}{2}[2 \cdot \frac{1}{xy}+(xy-1)\frac{1}{xy}]\\~~~~~~=\frac 12[2\cdot 1+(xy-1)\cdot 1]\\~~~~~~=\frac 12(xy+1)\,\,\text{(showed)}$