Question

You plan to accumulate $450,000 over a period of 12 years by making equal annual deposits in an account that pays an annual interest rate of 9% (assume all payments will occur at the beginning of each year). What amount must you deposit each year to reach your goal?​

Answer 1

Answer:

He should pay $20,498(approx) each year to reach his goal.

Step-by-step explanation:

As per the question,

We need to find the amount that we pay annualy.

As we know,

Given:

Years (t) = 12

Interest rate (i) = 9

Rate in decimal(r) = 0.09

No. of compounds per period (n) = 1

Future value (FV) = 450,000

So, we need to find payement per period(PMT)

As we know the formula:

$FV = PMT (((1 + \frac{r}{n} )^{nt} - 1) \div (\frac{r}{n} )) \times (1+ \frac{r}{n} )\\450000=PMT(((1.09)^{12}-1) \div (0.09))\times (1.09)))\\450000=PMT((\frac{2.812-1}{0.09})\times1.09)\\ 450000=PMT(20.140\times 1.09)\\450000=PMT\times21.953\\PMT=\frac{450000}{21.953} \\PMT=20,498(approx)$

Hence,

He should pay $20,498(approx) each year to reach his goal.

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Answer 2

Answer:

He should spend $ 20,498 (approx) each year to achieve his goal.

Step-by-step explanation:

Given:

Years (t) = 12

Interest rate (i) = 9

Rate in decimal (r) = 0.09

To find: the amount that we spend annually.

Number of compounds per period (n) = 1

Future value (FV) = $ 450,000

So, we required to find payment per period (PMT)

As we know the formula:

$F V=P M T\left(\left(\left(1+\frac{r}{n}\right)^{n t}-1\right) \div\left(\frac{r}{n}\right)\right) \times\left(1+\frac{r}{n}\right)$

$\left.\left.450000=P M T\left(\left((1.09)^{12}-1\right) \div(0.09)\right) \times(1.09)\right)\right)$

$450000=P M T\left(\left(\frac{2.812-1}{0.09}\right) \times 1.09\right)$

$450000=P M T(20.140 \times 1.09)$

$450000=P M T \times 21.953$

$P M T=\frac{450000}{21.953}$

PMT = 20,498 (approx)

He should spend $ 20,498 (approx) each year to achieve his goal.

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