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A sum of rupees 100 is invested at 8% per annum Simple Interset. Calculate the interest at the end of $1_{st}, 2_{nd}, 3_{rd} ....$ years. Is the sequence of interest an Arithmetic Progression. Find the interest at the end of 30 years.

Hint: Arithmetic Progression follows the rule of $a_n = a+(n-1)d$ where $n$ the term of the AP

My Problem: I am having problem in creating the Arithmetic Progression in this Question, because In this I am totally confused about concept.

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This looks a lot like homework. If so, that's OK, but use the homework tag. Also, give us your thoughts on the problem so far, so we can provide appropriate hints. –  Brett Frankel Oct 30 '12 at 1:08
    
You should get used to the idea of doing your own homework, or at the very least posting what you have tried so far so we know what you are struggling with. –  Arthur Collé Oct 30 '12 at 1:08

2 Answers 2

up vote 1 down vote accepted

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 $100\cdot8\% + 100\cdot8\%$

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 $$a_n = a+n\cdot d$$ $$a_n=100+n\cdot8$$

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Hint: how much interest is there in one year? For simple interest, that doesn't add to the principal, so the interest in each year is the same.

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