A man invests 10000 for 3 years at a certain rate of interest, compounded annually. At the end of one year, it amounts to 11200. Calculate :
(i) the rate of interest per annum;
(ii) the interest accrued in the second year;
(ii) the amount at the end of the third year.​

Step-by-step explanation:

10000, After 1 year amount = 11200 so interest after 1 year = 1200

Amount = P(1+R/100)T

11200 = 10000(1+R/100)1

R = 12%

Interest for the second year = 11200(1+12/100)1 - 11200 = 1344

after 3 years = 10000(1+12/100)3 = 14049.28

translate k to n and reflect across the line containing jk.