Hey,
Merry Christmas!
Jane's mother puts $300 into an account at the *beginning of each year* to pay for
Jane's education in 5 years' time. if 6% p.a. is paid quarterly, how much money will
Jane's mother have at the end of the 5 years?
. So the solution is as follows:
A1: 300*(1+0.015)^4
A2: (A1+300)*(1+0.015)^4 = (300*(1.015)^4 +300)*(1.015)^4
A3: (A2+300)*(1+0.015)^4
A4: (A3+300)*(1+0.015)^4
A5: (A4+300)*(1+0.015)^4
Add them up and you get your 1799.79 answer.
so can I use the sum formula to add them up S_5= (a(r^n-1)/r-1 ????
Merry Christmas!
Jane's mother puts $300 into an account at the *beginning of each year* to pay for
Jane's education in 5 years' time. if 6% p.a. is paid quarterly, how much money will
Jane's mother have at the end of the 5 years?
. So the solution is as follows:
A1: 300*(1+0.015)^4
A2: (A1+300)*(1+0.015)^4 = (300*(1.015)^4 +300)*(1.015)^4
A3: (A2+300)*(1+0.015)^4
A4: (A3+300)*(1+0.015)^4
A5: (A4+300)*(1+0.015)^4
Add them up and you get your 1799.79 answer.
so can I use the sum formula to add them up S_5= (a(r^n-1)/r-1 ????
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yes, you can, but you must be careful in assigning values for a , r & n
here, r = 1.015^4, a = 300*1.015 , n = 5, so
S = 300*1.015^4((1.015)^20 - 1)/(1.015^4-1) = 1799.79 <--------
here, r = 1.015^4, a = 300*1.015 , n = 5, so
S = 300*1.015^4((1.015)^20 - 1)/(1.015^4-1) = 1799.79 <--------
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sorry, that was a typo.
in the actual calculation, you can see that "a" has been taken as 300*1.015^4
in the actual calculation, you can see that "a" has been taken as 300*1.015^4
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Why, NO! I don't want to do this crap when I reach higher in class hey what subject is this tell me so i can cross it off my list