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Irene deposited \$200 in a bank on 1st January and on the first day of each of the following months. At the end of June, when the interest was calculated, he found that his account balance was \$1206.30. Find the rate of a simple interest paid by the bank?

$S_n = \frac{n(A_1+A_n)}{2}\\ S_n = 1206.30\\ A_1=200\\ A_n=\;?\\ n=6$

How to calculate the interest paid by the bank?

konewka
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user184563
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1 Answers1

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$A_n = A_0 + ni$, so $i = \frac{A_6-A_0}{6}$, as we have simple interest here.

Andrei Rykhalski
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  • Sn = [(A1+An)n]/2 Sn = 1206.30 A1=200 An=202.10 n=6

    ====> i = (202.10 - 200)/6 = 0.35

    BUT ANSWER 1.8% ????

     – user184563 Oct 15 '14 at 15:25
  • answer given is 1.8%, pls help to show how to derive the answer?? – user184563 Oct 15 '14 at 15:53
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    Ok, sorry, this problem is a bit harder that I thought, I missed an important statement due to type-setting. However, I still don't understand the condition. Irene deposited $6200 = 1200$$ and has $1206.30$$. What interest rate are we looking for? Monthly? Annual? Total for period? As it's a simple interest, we have $+200$ dollars at the start of each month and $+200i$ dollars at the end of each month with the total of $1206.30$, thus $1206.30 = 6 * 200 + 6 * i \Rightarrow i = 1.05$. Is this understanding of the problem correct? – Andrei Rykhalski Oct 15 '14 at 15:53
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    So please check all numbers and condition of the problem. I have some assumptions but they are related to your check. – Andrei Rykhalski Oct 15 '14 at 15:55
  • the interest rate is calculated 6 months later, as stated in question "At the end of June, when the interest was calculated".

    I supported your answer ~ 1.05%

     – user184563 Oct 16 '14 at 05:53