How to calculate the compound interest (compounded every 3 months) withh annual bonus if the balance is more than a certain amount.

Find the savings growth for a scheme whereby you are paid annual interest of 2.45%, compounded every 3 months, with an annual bonus of £40 if the total invested is over £3,000. The initial investment is £2500 and is invested for 12 years. What are the final savings (to 2 decimal places)?

Here is my code:

   x=input('initial amount    ');

   p=input('number of years   ');

   m=p*4; %%dealing with 3 months period instead of years 
       %%for easier compound interest calculations

    for k=1:m;

    intereset_every_3_month=x*(0.6125/100);
    x=intereset_every_3_month+x;
     if k==4 || k==8 || k==12 || k==16 || k==20 || k==24 || k==28 || k==32 || k==36 || k==40 || k==44 || k==48
     if x>3000
      x=x+40;
     end
    end

      disp(round(x));
     disp(intereset_every_3_month);
end

The conditional using the or operator is to ensure that the increase is only applied annually (every 4*3 months period)

My answer is 3562, however the correct answer is 3517. Am I doing something wrong here?

[dpb]

Crudely (ignoring the partial year in which the 3000 threshold was crossed) I got 3521 so since the desired answer is given as 3517 I'd suspect it's correct.

I'd rewrite on r=r_input/4; and loop until total >3000, then use the additive term.

Alternatively, being an engineer and if only looking to get the right answer, solve for M=month_exceed_3k, then use the power formula

t=t*(1+r)^M; % first M months
t=(t+40)^(48-M); % remaining

:)