A deposit of 10,000 is done. During first year, the bank credits an annual effective interest rate of $\text {i}$. During the second year, the bank credits an annual effective interest rate of $\text {i-5%}$. After two years, he has 12,093.75
What would be the amount earned by an annual effective interest rate of $\text {i+9%}$ for each of the three years?
So, we need to solve for $\text {i}$ from $A(2)=10,000[1+(i-0.05)]^2=12,093.75$
$$i=\Bigg(\frac{12,093.75}{10,000}\Bigg)^{1/2}-0.95, \text {for t=2}$$
But when substituted in $A(2)$, $A(2)\ne 12,093.75$
I'm new at this. How it is supposed to be done?
If the first year's effective interest rate is $i$, then at the end of the first year, the deposit will be $10000(1+i)$. Then, if the second year's rate is $i - 0.05$, then at the end of the second year, the deposit will be $$10000(1+i)(1 + i - 0.05).$$ This is the amount that must equal $12093.75$. Then the question asks, if the interest rate had been $i+0.09$ for all three years, then I would interpret that to mean the accumulated value $$10000(1 + i + 0.09)^3,$$ where $i$ was the rate calculated earlier.