Differentiation (Part-38) | S N De

$\,26.\,$  An A.P. consists of $\,n\,$ terms. If the sum of its first three terms is $\,x\,$ and the sum of the last three terms is $\,y\,$, then show that, the sum of all the terms of the A.P. is $\,~~\frac n6(x+y).$

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

So, according to the problem,

$\,\,a+(a+d)+(a+2d)=x \\ \Rightarrow 3(a+d)=x \\ \Rightarrow a+d=\frac x3 \rightarrow(1)$

Again, $\,l+(l-d)+(l-2d)=y \\ \Rightarrow 3(l-d)=\frac y3 \\ \Rightarrow l-d=\frac y3 \rightarrow(2)$

From $\,(1), (2),\,$ we get, 

$\,(a+d)+(l-d)=\frac x3+\frac y3 \\ \Rightarrow a+l=\frac 13(x+y)$

$\,\therefore \,$ the sum of all the terms of the A.P. is 

$=\frac n2 (\text{first term + last term})\\=\frac n2\times \frac 13(x+y)\\=\frac n6(x+y).$

$\,27.\,$ A farmer undertakes to pay off a debt of Rs. $\,2700\,$ by monthly instalments. He pays Rs. $\,200\,$ as the first instalment and increases every subsequent instalment by Rs. $\,25\,$ over the immediate previous instalment. In how many instalments his debt will be cleared up?

Sol. According to the problem, $\,a=200,\,d=25\,\,$ where $\,a=$ first term and $\,d\,$ being the common difference.

So, $\,S_n=200+225+250+\cdots \text{to n terms}\\~~~~~=\frac n2[2\times a+(n-1)d]\\~~~~~=\frac n2[2\times 200+(n-1)\times 25]\\~~~~~=\frac n2(375+25n) \\ \Rightarrow 2700=\frac n2 \times 25(n+15) \\ \Rightarrow \frac{2700}{25}=\frac n2(n+15) \\ \Rightarrow 108\times 2=n^2+15n  \\ \Rightarrow n^2+15n-216=0  \\ \Rightarrow  n=\frac{-15 \pm\sqrt{15^2-4 \times 1\times (-216)}}{2}\\~~~~~~~=\frac{-15 \pm\sqrt{225+864}}{2}\\~~~~~~~=\frac{-15 \pm\sqrt{1089}}{2}\\~~~~~~~=\frac{-15 \pm33}{2}\\ \Rightarrow n=\frac{-15+33}{2}=9,\\ ~~ n=\frac{-15-33}{2}=-24.$

But $\,\,n \,$ can not be $\,-24,\,$ and so, in $\,9\,$ instalments his debt will be cleared up.

$\,28.\,~~$A person pays Rs. $\,975\,$ in monthly instalments, each instalment is less than the former by Rs. $\,5\,$. The amount of first instalment is Rs. $\,100\,$. In what time will the entire amount to be paid ?

Sol. According to the problem, $\,a=100,\,d=(-5)\,\,$ where $\,a=$ first term and $\,d\,$ being the common difference.

So, $\,S_n=100+95+90+\cdots \text{to n terms}\\~~~~~=\frac n2[2\times a+(n-1)d]\\~~~~~=\frac n2[2\times 100+(n-1)\times (-5)]\\~~~~~=\frac n2(205-5n) \\ \Rightarrow 975=\frac n2 \times 5(41-n) \\ \Rightarrow \frac{975 \times 2}{5}=41n-n^2 \\ \Rightarrow n^2-41n+390=0 \\ \Rightarrow n=\frac{-(-41)\pm\sqrt{41^2-4 \times 1\times 390}}{2 \times 1}\\ \Rightarrow n=\frac{41\pm\sqrt{1681- 1560}}{2}\\ \Rightarrow n=\frac{41\pm\sqrt{121}}{2}\\ \Rightarrow n=\frac{41 \pm 11}{2} \\ \Rightarrow n=26,15.$

Since min $\{15,26\}=15\,$, so we can say that in $\,15\,$ months   the entire amount will be paid .

$\,29.\,$ If $\,1\,$ paisa is saved today, $\,2\,$ paise next day, $\,3\,$ paise succeeding day and so on, what will be the total savings in $\,365\,$ days?

Sol. According to the problem, let the total savings in $\,365\,$ days be denoted by $\,S.$

So, $\,S=1+2+3+\cdots+ 365\\~~~=\frac{365}{2}[2 \times 1+(365-1)\times 1] \,\,[**]\\~~=\frac{365}{2} \times [2+364]\\~~=\frac{365}{2} \times 366\\~~=66795.$

Hence, $\,66795\,$ paisa or $\,667.95\,$ rupees will be the total savings in $\,365\,$ days

Note[**] : Here , $\,S=\frac n2[2a+(n-1)d],$ where $\,a\,,d\,$ denote first term of the series and common difference respectively.

$\,30.\,$ The cost of sinking a tubewell is $\,25\,$ paise for the first metre and an additional $\,4\,$ paise for every subsequent metres. Find the cost of last metre and the total cost of sinking the tubewell $\,500\,$ metres deep.

Sol. The cost of sinking a tubewell is $\,25\,$ paise for the first metre, $\,25+4=29\,$ paisa for the $\,2\,$nd metre and so on.

Let $\,S\,$ denotes the total cost of sinking the tubewell $\,500\,$ metres deep.

So, $\,S=25+29+33+\cdots \text{to 500 terms}\\~~~=\frac{500}{2}[2\times 25+(500-1)\times 4]\\~~~=250\times 2046 \\~~~=511500\rightarrow(1)$

Hence, from $\,(1),\,$ we can say that the total cost of sinking the tubewell $\,500\,$ metres deep is $\,\,511500\,\,$ paisa or $\,5115\,$ rupees. 

Let $\,t_{500}\,\,$ denotes the cost of last metre  so that 

$\,t_{500}=25+(500-1)\times 4\,\,[*]\\~~~~~~=2021\rightarrow(2)$

Hence, the cost of last metre is : $\,2021\,$ paisa or $\,20.21\,$ rupees.

Note[*] : We know that $\,n\,$-th term is denoted by $\,t_n\,$ and is given by $\,t_n=a+(n-1)\times d,\,\,$ where $\,a\,$ is the first term and $\,d\,$ being the common difference of the Arithmetic Progression.

$\,31.\,$  The rate of monthly salary of an office assistant is increased annually in A.P. If he was drawing Rs. $\,2000\,$ a month during the $\,11\,$ th year and Rs. $\,3800\,$ a month during $\,29\,$ th year, find out his initial salary and rate of annual increment. Find out also his salary at the time of retirement on completion of $\,32\,$ years of service.

Sol.  Let the initial salary be $\,a\,$ and annual increment be $\,d.$ 

So, $\,t_n=a+(n-1)\times d\\ \Rightarrow t_{11}=a+(11-1)d \\ \Rightarrow 2000=a+10d \rightarrow(1) \\ \text{and}\,~~ 3800=a+(29-1)d \\ \Rightarrow 3800=a+28d \rightarrow(2)$

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

$\,18d=3800-2000 \\ \Rightarrow d=\frac{1800}{18}=100.$ 

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

$\,2000=a+10\times 100 \\ \Rightarrow a=2000-1000 \\ \therefore \, a=1000.$

Hence, the initial salary is $\,1000\,$ rs.  and annual increment is $\,100.$ 

So, the salary at the end of $\,32\,$ years is :

$\,=a+(n-1)\times d\\=1000+(32-1)\times 100\\=1000+3100\\=4100.$

$\,\therefore \,\,$ His salary at the time of retirement on completion of $\,32\,$ years of service is : $\,4100\,$ rs.

$\,32.\,$ The angles of a polygon are A.P. having common difference $\,5^{\circ}\,$. If the least angle be $\,120^{\circ},\,$ find the number of sides of the polygon.

Sol. For a polygon with $\,n\,$ sides, the sum of interior angles 

$=(n-2) \times 180^{\circ}\rightarrow(1)$

Again, the sum of interior angles can be calculated by using the formula,

$\,\frac n2[2a+(n-1)\times d]\\=\frac n2[2\times 120^{\circ}+(n-1)\times 5^{\circ}]\rightarrow(2)$

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

$\,\,(n-2)\times 180=\frac n2[240+5(n-1)] \\ \Rightarrow 180n-360=\frac n2[240+5n-5]\\ \Rightarrow 2(180n-360)=n(235+5n) \\ \Rightarrow 360n-720=235n+5n^2 \\ \Rightarrow 0=5n^2+235n-360n+720 \\ \Rightarrow 0=5n^2-125n+720 \\ \Rightarrow 5(n^2-25n+144)=0 \\ \Rightarrow n^2-9n-16n+144=0 \\ \Rightarrow n(n-9)-16(n-9)=0 \\ \Rightarrow (n-9)(n-16)=0 \\ \Rightarrow n=9,16.$

So, the number of the sides of the polygon is $\,9\,$ or $\,16.$