Question

Use) `A = P_1(1+r)^2 + P_2(1+r)`?

A firm deposit 3200 into savings account for 2 years, at the begging of second year, an additional 1800 is deposited. If a total of 5207 is in the account at the end, what is the annual interest rate.

answer should be 2.5%

Answer 1

This equation is a quadratic in 1+r

Explanation:

Make the substitute `x=1+r` and you'll see.

`0=P_1(1+r)^2+P_2(1+r)-A`

`0=P_1x^2+P_2x-A`

I'll just quickly use the quadratic formula rather than solving for x step by step.

`x=(-P_2+-sqrt(P_2^2+4P_1A))/(2P_1)`

`1+r=(-P_2+-sqrt(P_2^2+4P_1A))/(2P_1)`

`r=(-P_2+sqrt(P_2^2+4P_1A))/(2P_1)-1`

Plug in your numbers
`P_1=3200, P_2=1800, A=5207`
And the result is 0.025, which if we say `100%=1, %=1/100`, then we get the result of
`2.5 1/100=2.5%`