Let's say you have 100 rupees at 8% a year:
- At the end of the first year, the total interest is $100\cdot8\%$.
- At the end of the second year, the interest is $100\cdot8\% +
So, after $n$ years, you'll have a total interest owed as $n\cdot(100\cdot8\%)$.
Now, think about how much you would need to pay each year.
- At the end of the first year, you'd pay $100 + 100\cdot8\%$ rupees.
- At the end of the second year, you'd pay $100 + 100\cdot8\% + 100\cdot8\%$
Thus, at the end of the $n$th year, you'd pay $100 + 100\cdot8\%$ rupees.
$a$ is the starting value of the sequence, so $a=100$
$d$ is the "common difference" of the sequence--that's how much each term is greater than the last term. Thus, $d = 100\cdot8\%=8$.
Thus, $$a_n=a + (n - 1)\cdot d$$
However, there is an offset--the sequence starts at 100, then 108, then 116, etc. We want it to count 108, 116, 124, etc. So, our altered series is $$a_n = a+n\cdot d$$