Chapter 11 - Sequences; Induction; the Binomial Theorem - Section 11.3 Geometric Sequences; Geometric Series - 11.3 Assess Your Understanding - Page 845: 93

Around $\$305.10$ must be placed in the savings account each month.

Let us consider that $P$ is defined as the deposit in dollars made at the end of each payment period for annuity, when a person pays $i$ percent interest per payment period. The formula for amount $A$ of the annuity after $n$ deposits can be written as: $ A=P\cdot \dfrac{(1+i)^{n}-1}{i}$ We are given that $A=50,000 \\ n=(12)(10)=120 \\ i=\dfrac{0.06}{12}=0.005$ Substitute the given values into the formula above to obtain: $50,000=P \cdot \dfrac{(1+0.005)^{120}-1}{0.005}$ Isolate $P$ by multiplying $\dfrac{0.005}{(1+0.005)^{120}-1}$ to both sides: \begin{align*}50,000 \cdot \dfrac{0.005}{(1+0.005)^{120}-1}&=P \cdot \dfrac{(1+0.005)^{120}-1}{0.005} \cdot \dfrac{0.005}{(1+0.005)^{120}-1}\\ \\\dfrac{50,000 \cdot (0.005)}{(1+0.05)^{120}-1}&=P\\ \\305.10&\approx P\end{align*}