I am struggling to apply the arithmetic series formula in solving this word problem below, and any help would be appreciated:

The Problem:

Hercy wants to save money for his first car. He puts money in the bank every day.

He starts by putting in 1 dollar on Monday, the first day. Every day from Tuesday to Sunday, he will put in 1 dollar more than the day before. On every subsequent Monday, he will put in 1 dollar more than the previous Monday. Given an arbitrary n, write a general formula that expresses the total amount of money he will have in the bank at the end of the nth day in terms of n. (Hint: use an arithmetic progression)

My ideas so far:

I see this as summing $(1+2+...+7) + (2 +3+..+8) + (3+4+..+9) + ... + \text{remaining days}$

Which can be thought of as $\sum_{i=1}^7 i + 7(0) + \sum_{i=1}^7 i + 7(1) + \sum_{i=1}^7 i + 7(2) $ + ... +

I can get the current week # by doing $\lfloor((n-1)/7)\rfloor + 1$

I can get the current day# by doing $n-1 \pmod 7 + 1$.

So let's say n = 20., that is, 2 weeks and 6 days. I can express n with this formula:

$\sum_{i=1}^7 i + 7(0) + \sum_{i=1}^7 i + 7(1) + \sum_{i=1}^6 i + 7(2)$

but i'm struggling to get more general than that, and I'm not even sure if this is the correct way.


2 Answers 2


A whole week is $\$28$ plus $\$7$ more every time (every week starts with $\$7(w-1)+1$). Hence after $w$ weeks, the total is


Then if $d$ days remain, sum from $7(w-1)+1$ to $7(w-1)+d$, which is $$7(w-1)d+\dfrac{d(d+1)}2=\dfrac{d(14w+d-13)}2=\dfrac{(n-7w)(n+7w-13)}2.$$

You get $w,d$ from $n=7w+d$.


You have almost solved the problem. I'll give you a couple of hints, that should be enough. Let $n^{th}$ day$=$ $a^{th}$ week $+$ $b^{th}$ day. You already know $a$ in terms of $n$, so you know $b$ in terms of $n$ too. Since you know $a$, you know how much money Hency deposited in the incomplete week, because you know how much he deposited on the Monday of that week. For the complete weeks you have already obtained a summation, just evaluate it to get a closed form.


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