A carpenter was hired to build 192 window frames. The first day he made five frames and each day, thereafter he made two more frames than he made the day before. How many days did it take him to finish the job?
Here, $a=5$ and $d=2$
Let he finished the job in $n$ days.
Then,
$$\begin{aligned} S_n & =192 \\ S_n & =\frac{n}{2}[2 a+(n-1) d] \\ \Rightarrow \quad192 & =\frac{n}{2}[2 \times 5+(n-1) 2] \\ \Rightarrow \quad192 & =\frac{n}{2}[10+2 n-2] \end{aligned}$$
$$\begin{aligned} \Rightarrow \quad & 192 =\frac{n}{2}[8+2 n] \\ \Rightarrow \quad & 192 =4 n+n^2 \\ \Rightarrow \quad & n^2+4 n-192 =0 \\ \Rightarrow \quad & (n-12)(n+16)=0 \\ \Rightarrow \quad & n =12,-16 \\ \therefore \quad & n =12 \quad [\because n\ne -16] \end{aligned}$$
The sum of interior angles of a triangle is $180^{\circ}$. Show that the sum of the interior angles of polygons with $3,4,5,6, \ldots$ sides form an arithmetic progression. Find the sum of the interior angles for a 21 sided polygon.
We know that, sum of interior angles of a polygon of side $n=(2 n-4) \times 90 \Upsilon=(n-2) \times 180 \Upsilon$
Sum of interior angles of a polygon with sides 3 is 180 .
Sum of interior angles of polygon with side $4=(4-2) \times 180 \Upsilon=360 \Upsilon$
Similarly, sum of interior angles of polygon with side $5,6,7 \ldots$ are $540^{\circ}, 720^{\circ}, 900^{\circ}, \ldots$
The series will be $180^{\circ}, 360^{\circ} 540^{\circ}, 720^{\circ}, 900^{\circ}, \ldots$
Here, $\quad a=180 \Upsilon$
and $\quad d=360^{\circ}-180^{\circ}=180^{\circ}$
Since, common difference is same between two consecutive terms of the series.
So, it form an AP.
We have to find the sum of interior angles of a 21 sides polygon.
It means, we have to find the 19th term of the above series.
$$\begin{aligned} \therefore \quad a_{19} & =a+(19-1) d \\ & =180+18 \times 180=3420 \end{aligned}$$
A side of an equilateral triangle is 20 cm long. A second equilateral triangle is inscribed in it by joining the mid-points of the sides of the first triangle. The process is continued as shown in the accompanying diagram. Find the perimeter of the sixth inscribed equilateral triangle.
Side of equilateral $\triangle A B C=20 \mathrm{~cm}$. By joining the mid-points of this triangle, we get another equilateral triangle of side equal to half of the length of side of $\triangle A B C$.
Continuing in this way, we get a set of equilateral triangles with side equal to half of the side of the previous triangle.
$$\begin{aligned} \therefore \quad \text { Perimeter of first triangle } & =20 \times 3=60 \mathrm{~cm} \\ \text { Perimeter of second triangle } & =10 \times 3=30 \mathrm{~cm} \\ \text { Perimeter of third triangle } & =5 \times 3=15 \mathrm{~cm} \end{aligned}$$
Now, the series will be $60,30,15, \ldots$
Here,
$$\begin{aligned} & a=60 \\ & r=\frac{30}{60}=\frac{1}{2} \quad\left[\because \frac{\text { second term }}{\text { first term }}=r\right] \end{aligned}$$
We have, to find perimeter of sixth inscribed triangle. It is the sixth term of the series.
$$\begin{aligned} \therefore \quad a_6 & =a r^{6-1} \quad \left[\because a_n=a r^{n-1}\right]\\ & =60 \times\left(\frac{1}{2}\right)^5=\frac{60}{32}=\frac{15}{8} \mathrm{~cm} \end{aligned}$$
In a potato race 20 potatoes are placed in a line at intervals of 4 m with the first potato 24 m from the starting point. A contestant is required to bring the potatoes back to the starting place one at a time. How far would he run in bringing back all the potatoes?
According to the given information, we have following diagram.
Distance travelled to bring first potato $=24+24=2 \times 24=48 \mathrm{~m}$
Distance travelled to bring second potato $=2(24+4)=2 \times 28=56 \mathrm{~m}$
Distance travelled to bring third potato $=2(24+4+4)=2 \times 32=64 \mathrm{~m}$
Then, the series of distances are $48,56,64, \ldots$
Here,
$$\begin{aligned} & a=48 \\ & d=56-48=8 \\ \text{and}\quad & n=20 \end{aligned}$$
To find the total distance that he run in bringing back all potatoes, we have to find the sum of 20 terms of the above series.
$$\begin{aligned} \therefore \quad S_{20} & =\frac{20}{2}[2 \times 48+19 \times 8] \quad\left[\because S_n=\frac{n}{2}\{2 a+(n-1) d\}\right] \\ & =10[96+152] \\ & =10 \times 248=2480 \mathrm{~m} \end{aligned}$$
In a cricket tournament 16 school teams participated. A sum of ₹ 8000 is to be awarded among themselves as prize money. If the last placed team is awarded ₹ 275 in prize money and the award increases by the same amount for successive finishing places, how much amount will the first place team receive?
Let the first place team got ₹ $a$.
Since, award money increases by the same amount for successive finishing places.
Therefore series is an AP.
Let the constant amount be $d$.
$$\begin{aligned} &\begin{array}{ll} \text { Here, } & l=275, n=16 \text { and } S_{16}=8000 \\ \therefore & l=a+(n-) d \\ \Rightarrow & l=a+(16-1)(-d) \end{array}\\ &\text { [we take common difference ($-$ve) because series is decreasing] } \end{aligned}$$
$$\begin{array}{ll} \Rightarrow & 275=a-15 d \quad \text{... (i)}\\ \text { and } & S_{16}=\frac{16}{2}[2 a+(n-1) \cdot(-d)] \\ \Rightarrow & 8000=8[2 a+(16-1)(-d)] \\ \Rightarrow & 8000=8[2 a-15 d] \\ \Rightarrow & 1000=2 a-15 d \quad \text{... (ii)} \end{array}$$
On subtracting Eq. (i) from Eq. (ii), we get
$$\begin{aligned} & & (2 a-15 d)-(a-15 d) & =1000-275 \\ \Rightarrow & & 2 a-15 d-a+15 d & =725 \\ \therefore & & a & =725 \end{aligned}$$
Hence, first place team receive ₹ $725$.