Question #280577

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? 


1
Expert's answer
2021-12-17T13:39:55-0500

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+0.0612)1+...+100(1+0.0612)60A=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+0.0612)1+100(1+0.0612)2+...+100(1+0.0612)60A=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,

a1=100(1+0.0612)r=(1+0.0612)n=60a_1=100(1+\frac{0.06}{12}) \\ r=(1 +\frac{0.06}{12}) \\ n=60

Now, we can apply the formula for the sum of a finite geometric sequence,

A=Sn=a(1rn1r)=100(1+0.0612)(11.0056011.005)=7011.89A=S_n \\ =a(\frac{1-r^n}{1-r}) \\ =100(1+\frac{0.06}{12})(\frac{1-1.005^{60}}{1-1.005}) \\ = 7011.89

Therefore the balance A in account at the end of 5 years is 7011.89


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