The first term of an AP is a and the sum of the first $p$ terms is zero, show that the sum of its next $q$ terms is $\frac{-a(p+q) q}{p-1}$.
Let the common difference of an AP is $d$.
According to the question,
$$\begin{aligned} & \qquad S_p=0 \\ & \Rightarrow \quad \frac{p}{2}[2 a+(p-1) d]=0 \quad {\left[\because S_n=\frac{n}{2}\{2 a+(n-1) d\}\right]}\\ & \Rightarrow \quad 2 a+(p-1) d=0 \\ & \therefore \quad d=\frac{-2 a}{p-1} \end{aligned}$$
$$\begin{aligned} &\text { Now, sum of next } q \text { terms }\\ &\begin{aligned} S & =S_{p+q}-S_p=S_{p+q}-0 \\ & =\frac{p+q}{2}[2 a+(p+q-1) d] \\ & =\frac{p+q}{2}[2 a+(p-1) d+q d] \\ & =\frac{p+q}{2}\left[2 a+(p-1) \cdot \frac{-2 a}{p-1}+\frac{q(-2 a)}{p-1}\right] \\ & =\frac{p+q}{2}\left[2 a+(-2 a)-\frac{2 a q}{p-1}\right] \\ & =\frac{p+q}{2}\left[\frac{-2 a q}{p-1}\right] \\ & =\frac{-a(p+q) q}{(p-1)} \end{aligned} \end{aligned}$$
A man saved ₹ 66000 in 20 yr. In each succeeding year after the first year, he saved ₹ 200 more than what he saved in the previous year. How much did he save in the first year?
Let saved in first year ₹ $a$. Since, each succeeding year an increment ₹ 200 has made.
So, it forms an AP whose
First term $=a$, common difference $(d)=200$ and $n=20 \mathrm{yr}$
$$\begin{array}{lll} \therefore & S_{20}=\frac{20}{2}[2 a+(20-1) d] & {\left[\because S_n=\frac{n}{2}\{2 a+(n-1) d\}\right]} \\ \Rightarrow & 66000=10[2 a+19 d] \\ \Rightarrow & 66000=20 a+190 d \\ \Rightarrow & 66000=20 a+190 \times 200 \\ \Rightarrow & 20 a=66000-38000 \\ \Rightarrow & 20 a=28000 \\ \therefore & a=\frac{28000}{20}=1400 \end{array}$$
Hence, he saved ₹ 1400 in the first year.
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,
$$\begin{aligned} S_n & =192 \\ S_n & =\frac{n}{2}[2 a+(n-1) d] \\ \Rightarrow \quad192 & =\frac{n}{2}[2 \times 5+(n-1) 2] \\ \Rightarrow \quad192 & =\frac{n}{2}[10+2 n-2] \end{aligned}$$
$$\begin{aligned} \Rightarrow \quad & 192 =\frac{n}{2}[8+2 n] \\ \Rightarrow \quad & 192 =4 n+n^2 \\ \Rightarrow \quad & n^2+4 n-192 =0 \\ \Rightarrow \quad & (n-12)(n+16)=0 \\ \Rightarrow \quad & n =12,-16 \\ \therefore \quad & n =12 \quad [\because n\ne -16] \end{aligned}$$