A man accepts a position with an initial salary of ₹ 5200 per month. It is understood that he will receive an automatic increase of ₹ 320 in the very next month and each month thereafter.
(i) Find his salary for the tenth month.
(ii) What is his total earnings during the first year?
Since, the man get a fixed increment of ₹ 320 each month.
Therefore, this forms an AP whose First term $=5200$ and Common difference $(d)=320$
(i) Salary for tenth month i.e., for $n=10$,
$$\begin{array}{ll} & a_{10}=a+(n-1) d \\ \Rightarrow & a_{10}=5200+(10-1) \times 320 \\ \Rightarrow & a_{10}=5200+9 \times 320 \\ \therefore & a_{10}=5200+2880 \\ \therefore & a_{10}=8080 \end{array}$$
(ii) Total earning during the first year.
In a year there are 12 month i.e., $n=12$,
$$\begin{aligned} S_{12} & =\frac{12}{2}[2 \times 5200+(12-1) 320] \\ & =6[10400+11 \times 320] \\ & =6[10400+3520]=6 \times 13920=83520 \end{aligned}$$
If the $p$ th and $q$ th terms of a GP are $q$ and $p$ respectively, then show that its $(p+q)$ th term is $\left(\frac{q^p}{p^q}\right)^{\frac{1}{p-q}}$.
Let the first term and common ratio of GP be $a$ and $r$, respectively.
According to the question, $p$ th term $=q$
$$\begin{array}{lrl} \Rightarrow & a \cdot r^{p-1} & =q \quad \text{... (i)}\\ \text { and } & q \text { th term } & =p \\ \Rightarrow & a r^{q-1} & =p\quad \text{..... (ii)} \end{array}$$
On dividing Eq. (i) by Eq. (ii), we get
$$\begin{aligned} \frac{a r^{p-1}}{a r^{q-1}} & =\frac{q}{p} \\ \Rightarrow \quad r^{p-1-q+1} & =\frac{q}{p} \\ \Rightarrow \quad r^{p-q} & =\frac{q}{p} \Rightarrow r=\left(\frac{q}{p}\right)^{\frac{1}{p-q}} \end{aligned}$$
On substituting the value of $r$ in Eq. (i), we get
$a\left(\frac{q}{p}\right)^{\frac{p-1}{p-q}}=q \Rightarrow a=\frac{q}{\left(\frac{q}{p}\right)^{\frac{p-1}{p-q}}}=q \cdot\left(\frac{p}{q}\right)^{\frac{p-1}{p-q}}$
$$\begin{aligned} \therefore \quad(p+q) \text { th term } & T_{p+q}=a \cdot r^{p+q-1}=q \cdot\left(\frac{p}{q}\right)^{\frac{p-1}{p-q}} \cdot(r)^{p+q-1} \\ & =q \cdot\left(\frac{p}{q}\right)^{\frac{p-1}{p-q}}\left[\left(\frac{q}{p}\right)^{\frac{1}{p-q}}\right]^{p+q-1}=q \cdot\left(\frac{p}{q}\right)^{\frac{p-1}{p-q}}\left(\frac{q}{p}\right)^{\frac{p+q-1}{p-q}} \\ & =q \cdot\left(\frac{p}{q}\right)^{\frac{p-1}{p-q}}\left(\frac{p}{q}\right)^{\frac{-(p+q-1)}{p-q}}=q \cdot\left(\frac{p}{q}\right)^{\frac{p-1}{p-q}-\frac{(p+q-1)}{p-q}} \\ & =q \cdot\left(\frac{p}{q}\right)^{\frac{p-1-p-q+1}{p-q}}=q \cdot\left(\frac{p}{q}\right)^{\frac{-q}{p-q}} \\ a & =q \cdot\left(\frac{p}{q}\right)^{\frac{p-1}{p-q}} \end{aligned}$$
$$\begin{aligned} &\text { Now, }(p+q) \text { th term i.e., } a_{p+q}=a r^{p+q-1}\\ &\begin{aligned} & =q \cdot\left(\frac{p}{q}\right)^{\frac{p-1}{p-q}} \cdot\left(\frac{q}{p}\right)^{\frac{p+q-1}{p-q}} \\ & =q \cdot \frac{q^{\frac{p+q-1-p+1}{p-q}}}{p^{\frac{p+q-1-p+1}{p-q}}}=q \cdot\left(\frac{q^{\frac{q}{p-q}}}{\frac{q}{p^{p-q}}}\right) \end{aligned} \end{aligned}$$
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.
$$\begin{aligned} \therefore \quad a_6 & =a r^{6-1} \quad \left[\because a_n=a r^{n-1}\right]\\ & =60 \times\left(\frac{1}{2}\right)^5=\frac{60}{32}=\frac{15}{8} \mathrm{~cm} \end{aligned}$$