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 $S_{30}=1100$?
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:
This yields the equation $$780+10x+55\cdot4=1100$$ and solving we find that $x=10$.
Notice that we have not used the formula for arithmetic progressions here because there are few enough days between the 27th and 30th that we can list them individually. However, our solution is in the same spirit.