A deposit of N100 is made at the beginning of each month in an account that pays 6% interest, compounded monthly. The balance in the account at the end of 5 years is?
A deposit of $100 is made at the beginning of each month in an account that pays 6% interest, compounded monthly. The balance A in the account at the end of 5 years is
"A=100(1 +\\frac{0.06}{12})^1 +...+100(1 +\\frac{0.06}{12})^{60}"
The balance A in account at the end of 5 years is,
"A=100(1 +\\frac{0.06}{12})^1 + 100(1 + \\frac{0.06}{12})^2+...+100(1 +\\frac{0.06}{12})^{60}"
We can see that,
"a_1=100(1+\\frac{0.06}{12}) \\\\\n\nr=(1 +\\frac{0.06}{12}) \\\\\n\nn=60"
Now, we can apply the formula for the sum of a finite geometric sequence,
"A=S_n \\\\\n\n=a(\\frac{1-r^n}{1-r}) \\\\\n\n=100(1+\\frac{0.06}{12})(\\frac{1-1.005^{60}}{1-1.005}) \\\\\n\n= 7011.89"
Therefore the balance A in account at the end of 5 years is 7011.89
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