MHB Write down an equation & solve for x

[mathlearn]

Problem

A person starts saving money by depositing 5 coins in his till on the first day. After that , every day , he deposits 2 more than the amount she deposited in the till on the previous day

In order that the amount of money in the till at the end of 30 th day is 1100 coins, He deposits , from the 27th day $x$ rupees more than the amount deposited on the previous day.

Where do I need help

Write down an equation in $x$ and find the value of $x$ by solving it

Many Thanks :)

 

[MarkFL]

Let's begin by letting $D(n)$ be the amount deposited on the $n$th day. Now, we are told this amount increases linearly, and that $D(1)=5$. You are given the slope of this linear relationship, and you are given a point on that line...can you now use the point-slope formula to determine $D(n)$?

 

[mathlearn]

Let's begin by letting $D(n)$ be the amount deposited on the $n$th day. Now, we are told this amount increases linearly, and that $D(1)=5$. You are given the slope of this linear relationship, and you are given a point on that line...can you now use the point-slope formula to determine $D(n)$?

My Apologies I don't know how:)

 

[MarkFL]

Well, let's begin with the slope:

$$m=\frac{\Delta D}{\Delta n}=\frac{D(n+1)-D(n)}{(n+1)-n}=\frac{(D(n)+2)-D(n)}{(n+1)-n}=\frac{2}{1}=2$$

We could have simply observed that for each increase in the number of days by one day, the amount deposited increases by 2, but I wanted to show you the method when it's not as obvious.

Okay, we have a slope of 2 and the point $(n,D)=(1,5)$ and the point-slope formula then gives us:

$$D(n)-5=2(n-1)\implies D(n)=2n+3$$

So, now, we need to sum up the deposits for the first $n$ days to find out what the total is after those $n$ days.

$$S_{n}=\sum_{k=1}^{n}\left(2k+3\right)$$

We will need the following summation formulas:

$$\sum_{k=1}^{n}(a\cdot f(k))=a\cdot\sum_{k=1}^{n}(f(k))$$

$$\sum_{k=1}^{n}(f(k)\pm g(k))=\sum_{k=1}^{n}(f(k))\pm\sum_{k=1}^{n}(g(k))$$

$$\sum_{k=1}^{n}(1)=n$$

$$\sum_{k=1}^{n}(k)=\frac{n(n+1)}{2}$$

Now we may write:

$$S_{n}=2\sum_{k=1}^{n}\left(k\right)+3\sum_{k=1}^{n}(1)$$

Can you now apply the formulas above to write $S_{n}$ as a function of $n$?

 

[MarkFL]

You've indicate by PM a desire to use the formula for the sum of an AP:

$$S_n=\frac{n}{2}\left(2a_1+(n-1)d\right)$$

With $a_1=5$ and $d=2$, we obtain:

$$S_n=\frac{n}{2}\left(2\cdot5+2(n-1)\right)=n(5+n-1)=n(n+4)$$

Had we continued with the method I was using, we would find:

$$S_n=2\cdot\frac{n(n+1)}{2}+3n=n^2+n+3n=n^2+4n=n(n+4)$$

Okay, next we need to find $S_{27}$ so that we know how much was deposited after 27 days:

$$S_{27}=27(27+4)=27\cdot31=837$$

This means over the next 3 days, there needs to be $1100-837=263$ Rupees deposited. So, using the AP formula you cited, we have:

$$a_1=D(27)+x=2(27)+3+x=57+x$$

$$d=x$$

$$n=3$$

Hence:

$$S_3=\frac{3}{2}\left(2(x+57)+(3-1)x\right)=263$$

Can you now solve for $x$?

 

[mathlearn]

You've indicate by PM a desire to use the formula for the sum of an AP:

$$S_n=\frac{n}{2}\left(2a_1+(n-1)d\right)$$

With $a_1=5$ and $d=2$, we obtain:

$$S_n=\frac{n}{2}\left(2\cdot5+2(n-1)\right)=n(5+n-1)=n(n+4)$$

Had we continued with the method I was using, we would find:

$$S_n=2\cdot\frac{n(n+1)}{2}+3n=n^2+n+3n=n^2+4n=n(n+4)$$

Okay, next we need to find $S_{27}$ so that we know how much was deposited after 27 days:

$$S_{27}=27(27+4)=27\cdot31=837$$

This means over the next 3 days, there needs to be $1100-837=263$ Rupees deposited. So, using the AP formula you cited, we have:

$$a_1=D(27)+x=2(27)+3+x=57+x$$

$$d=x$$

$$n=3$$

Hence:

$$S_3=\frac{3}{2}\left(2(x+57)+(3-1)x\right)=263$$

Can you now solve for $x$?

Thank you very much :)

$$S_3=\frac{3}{2}\left(2x+114+3x-x\right)=263$$
$$S_3=\frac{3}{2}\left(4x+114 \right)=263$$
$$S_3=3 \left(4x+114 \right)=526$$
$$S_3= \left(12x+342\right)=526$$
$$S_3= 12x = 184$$
$$S_3= x = 15.3$$

Correct? :)

Many Thanks :)

 

[MarkFL]

I get:

$$x=\frac{46}{3}=15.\overline{3}$$

I was expecting an integral answer. I now see I misread the problem...what we want is 26 days first:

$$S_{26}=780$$

$$1100-780=220$$

$$a_1=D(26)+x=2(26)+3+x=55+x$$

And then:

$$S_4=\frac{4}{2}\left(2(x+55)+(4-1)x\right)=220$$

$$2(2(x+55)+3x)=220$$

$$2(2x+110+3x)=220$$

$$2(5x+110)=220$$

$$10(x+22)=220$$

$$x+22=22$$

$$x=0$$