-3(n + 3)=12 the solution is n =
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If Jerry contributes at the beginning of the month and withdraws at the end of the month, the final contribution earns 1 month's interest. The one before that earns 2 months' interest, so has a value of (1+0.017/12) times that of the last payment. In short, the sum is that of a geometric sequence with first term
a₁ = 300*(1+0.017/12)
and common ratio
r = 1+0.017/12
We assume Jerry contributes each month for 15 years, so a total of 180 payments. The sum is given by the formula for the sum of a geometric sequence.
Filling in your numbers, this is
If Jerry's contributions and withdrawal are at the end of the month, this balance is reduced by 1 month's interest, so is $61,460.
_____
We suppose the expected choice is $61,960. This supposition comes from the fact that a handwritten 4 is often confused with a handwritten 9. The usual simple calculation of future value uses end-of-the-month contributions by default. (a₁ = 300)
a₁ = 300*(1+0.017/12)
and common ratio
r = 1+0.017/12
We assume Jerry contributes each month for 15 years, so a total of 180 payments. The sum is given by the formula for the sum of a geometric sequence.
Filling in your numbers, this is
If Jerry's contributions and withdrawal are at the end of the month, this balance is reduced by 1 month's interest, so is $61,460.
_____
We suppose the expected choice is $61,960. This supposition comes from the fact that a handwritten 4 is often confused with a handwritten 9. The usual simple calculation of future value uses end-of-the-month contributions by default. (a₁ = 300)