anonymous
  • anonymous
Tian opened a saving account on January 1st with a deposite of $150. The following July 1st, January 1st, and July 1st, he made 3 more deposits of $150 each. The account paid interest at 3.75% per annum, compounded semi-annually. a) What amount did Tian have in his account at the end of the second year? b) How much interest did Tian earn by the end of the second year?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
Do it in segments remember that there are 12 months in a year. n=2
anonymous
  • anonymous
I tried R, refular deposits (payment) = 150, n, number of deposits or payment = 3, and i, interest rate per compounding period: 0.03765/2 = 0.01875. I got $458. 50 in the end using this formula: A = R [(1 + i)^n - 1 / i. However, the correct answers are: a) $617.09 and b) $17.09
anonymous
  • anonymous
n = 2 because it is compounded twice a year so every six months 3.75% of the money present in the account is added to the original amount invested.

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anonymous
  • anonymous
ohh.. so what's the function of the 'he made 3 more deposits of $150.'?
anonymous
  • anonymous
We know the original amount was compounded once before he made his second investment, then compounded again when he made his third investment than compounded again when he made his fourth investment
anonymous
  • anonymous
yup, he did.
anonymous
  • anonymous
To answer b) The first compound can be expressed as (0.0375)*150 + 150 The second compound he added 150 so for the second compound (first year we have the expression) ((((0.0375)*150 + 150) + 150)0.03 + ((((0.0375)*150 + 150) + 150)
anonymous
  • anonymous
I would probably just try answering this manually :L
anonymous
  • anonymous
ohhh, okay. So do I still use the formula or.. no?
anonymous
  • anonymous
well you could use it for every compound but make sure you change the principle accordingly
anonymous
  • anonymous
I'm sure there is a way to modify the equation so that ever time there is a compound 150 is added plus the interest
anonymous
  • anonymous
ohh.. okay. ohhhh.. like this: R + R(1 + i)^1 + R(1 + i)^2 + R(1+i)^3 +... + R (1+i)^n-1? And calculate the sum of the series: which is basically the formula?
anonymous
  • anonymous
oh, that one is for compounding interests isn't it?
anonymous
  • anonymous
yeah but I was wrong the formula has to take in account the added 150 after each compound

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