Question

A village is supplied water through pipe lines every third day. The water was first supplied on ${{1}^{st}}$ January. The last was supplied on ${{31}^{st}}$ January. How many times was water supplied in the month of January?
$10,000$ Kl of water supplied on ${{1}^{st}}$ January. Subsequently, the quota of water was increased by $500$ liters each time. How much water was supplied during the month of January?

Answer 1

To find the number of times water supplied, we will write down the some dates from starting and will observe that it will be in A.P. So, we will use the formula of ${n^{th}}$ term to find the number of times because the last date was given that would be the last term of the series.
For finding the total supplied, again we will write the quantity supplied for some days and will find out that it is also a series of A.P. Then, we will use the formula of sum of the A.P. and will simplify it to find the total supplied water.

Complete step by step answer:
Since, we have given the first date of supplied water that is $1$and the difference between two consecutive days that is $3$ because water is supplied on every third day from the previous day. And the last day of water supply is $31$. These all conditions denote that if we write the dates in a sequence, it will be a series of A.P. Since, we have some terms of A.P. So, we will use the formula of ${n^{th}}$ term to find the number of times of water supplied as:
$\Rightarrow {t_n} = a + \left( {n - 1} \right)d$
Where, $a$ is first date of water supplied, ${t_n}$ is the ${n^{th}}$ day of water supplied, $n$ is number of times or number of days of water supplied and $d$ is the difference between two consecutive days.
Now, we will substitute corresponding values in the formula as:
$\Rightarrow 31 = 1 + \left( {n - 1} \right)3$
Here, we will simplify the expression by using required calculations as:
$\Rightarrow 31 = 1 + 3n - 3$
$\Rightarrow 31 = 3n - 2$
$\Rightarrow 31 + 2 = 3n$
$\Rightarrow 33 = 3n$
Now, we can write the last step as:
$\Rightarrow n = \dfrac{{33}}{3}$
After simplification, we will have:
$\Rightarrow n = 11$
So, the water was supplied for $11$ days.
Now, we will look for the quantity of supplied water. Here, we have almost same condition as water supplied on first date is $10,000$ kl and the quantity of water is increasing subsequently that means difference of water quantity between two consecutive days that is $500$ letre or $0.5$ Kl and we got that the water was supplied $11$ days in January month. So, these all conditions denote that if we write the quantity of water in a sequence, it will be a series of A.P. Since, we have some terms of A.P. So, we will use the formula of sum of $n$ terms to find the number of total supplied water as:
$\Rightarrow S = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$
Where, $a$ is the quantity of water on the first date, $S$ is the total quantity of water supplied, $n$ is the number of times or number of days of water supplied and $d$ is the difference of quantity of water between two consecutive days.
Now, we will substitute the corresponding values in the formula and simplify it as:
$\Rightarrow S = \dfrac{{11}}{2}\left[ {2 \times 10,000 + \left( {11 - 1} \right) \times 0.5} \right]$
$\Rightarrow S = \dfrac{{11}}{2}\left[ {20,000 + 10 \times 0.5} \right]$
$\Rightarrow S = \dfrac{{11}}{2}\left[ {20,000 + 5} \right]$
$\Rightarrow S = \dfrac{{11}}{2}\left[ {20,005} \right]$
Now, we will substitute $5.5$ for $\dfrac{{11}}{2}$ using division and then will use multiplication to get the required answer as:
$\Rightarrow S = 5.5 \times 20,005$
$\Rightarrow S = 1,10,027.50$
Hence, the total number of times water supplied is $11$ and the total quantity of supplied water is $1,10,027.5$KL.

Note: When a series is written as:
$\Rightarrow a,a + d,a + 2d,a + 3d + ...$
The series is in A.P. Where, $a$ is the first term of the series and $d$ is the difference between two consecutive terms of the series that is a constant.