Anonymous
Anonymous asked in Science & MathematicsMathematics · 1 month ago

# Math Things: Johnny puts 25\$ at the end of each month for 4 years into his bank that pays...?

... 3.5% compounded monthly. Johnny makes no more deposits or withdrawals. determine the balance 10 years after the last deposit.

grade 11 math equation usually given would be

A= R (1+i)^n -1 divided by i

A= future value

R= present value (25)

Relevance

A = (R/i)[(1 + i)ⁿ - 1]

You mention in a previous comment that when you 'plug in values' you don't get the same values as me. I fear that your order of operations may be suspect.

We have that A(n) = (P/i)[(1 + i)ⁿ - 1]

Here, P is the monthly deposit, i is the monthly equivalent of the annual % rate and n is the number of depositing periods.

so, P = 25 and n = 4 x 12 => 48

Now, is the quoted figure of 3.5% p.a. an A.P.R or flat rate??

I will assume an A.P.R. as this is a standard quoted figure in the UK.

Now, 3.5% APR => (1.035)¹/¹² - 1

i.e. 0.002871 => 0.2871%/month....the value to use.

Note: 3.5% flat would be 3.5/12 => 0.2917/month

Then, B = (25/0.002871)[(1 + 0.002871)⁴⁸ - 1]

so, B = \$1284.64.....at end of 4 years

Now, we have interest compounded on this figure, monthly, for the next 10 years (120 months)

so, 1284.64(1.002871)¹²⁰ => \$1812.13

:)>

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• First, calculate what he has after the 4 years

A = (25 * (1 + 0.035/12)^(12 * 4) - 1) / (0.035/12)

A = 12 * (25 * (12.035/12)^(48) - 1) / 0.035

A = 12 * 1000 * (25 * (12.035/12)^(48) - 1) / 35

A = 12000 * (25 * (12.035/12)^(48) - 1) / 35

A = ‭1,286.0522641789499656143232475895‬

1286.05

Now apply the compound interest on 1286.05 for 10 years

1286.05 * (1 + 0.035/12)^(12 * 10) =>

1286.05 * (12.035/12)^120 =>

‭1,824.0623587020435210991373639548‬

1824.06

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• First consider the first \$25.  You can calculate the final value of that by

A = 25(1+0.035/12)^119

= 25(1.002917)^119 = 25(m)^119, m being the monthly interest rate

Now suppose at the end of the month, the \$25 was put into some other account, or even at another bank.  The value of that would be

25(m)^118

And the next month,

25(m)^117

At the end, you would add up all the amounts and get

25(m^119 + m^118 + ... + m^2 + m^1 + m^0)

That thing with all the powers in the parentheses is a geometric series.  By the formula for that sum (look up "sum of geometric series" for a refresher):

S = (1 - m^120) / (1 - m)

Can you take it from there?

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