# Coins in the till and arithmetic progressions

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

However, so that 1100 coins may be in the till on day 30, he deposits $$x$$ coins more than the amount he deposited on the previous day from day 27 onwards. What is $$x$$?

Use arithmetic progressions and the formula $$S_n=\frac{n}{2}\left(2a_1+(n-1)d\right)$$.

What I have tried: $$S_{26}=\frac{26} {2} \left\{2*5+(26-1)2 \right\}$$ $$S_{26}=13 \left\{10+50\right\}=780$$ The total in the till on the 30th day would then be $$S_{30}=\frac{30} {2} \left\{2*5+(30-1)2 \right\}$$ $$S_{30}=15 \left\{10+58 \right\}=1020$$ but isn't $$S_{30}=1100$$?

• Have you tried anything? Commented Oct 4, 2016 at 10:35
• @ParclyTaxel I have calculated the 26 & the 30 th terms Commented Oct 4, 2016 at 10:41

We know that 780 coins are in the till at the end of day 26. We can also work out how many coins were deposited on day 26: $5+2\cdot25=55$. Now we have another arithmetic progression:
• on day 27, $x+55$ coins are deposited
• on day 28, $2x+55$ coins are deposited
• on day 29, $3x+55$ coins are deposited
• and on day 30, $4x+55$ coins are deposited to bring the total to 1100.
This yields the equation $$780+10x+55\cdot4=1100$$ and solving we find that $x=10$.