COMPOUND INTEREST

COMPOUND INTEREST

1. A = P [ 1 + [ R/(100*n) n*t] ] P -----> Principle R -----> Rate % per annum n -----> No.of convertions per year T -----> Time in years 2.C.I=A-P      i.e., [ P [ 1+(R/(100*n)n*t] - 1 ] 3.When interest is calculated anually n=1, A = P[1+(R/100)t] 4.When time is in fraction, t = x * (1/y) year: A = P[1+(R/100)x] + [1 + (1/y)*R/100 ] 5.When rate od interest is R1% R2%  R3% for Ist year,IIndyear, IIIrd year respectively then amount, A = P [ 1 + (R1100) * [1+(R2/100)] [1+(R3/100)] 6.When difference between C.I and S.I on certain sum at rate% on Rs.x, [ C.I - S.I = sum * (r/100)2 ] i.e., [ D = P * (r/100)2 ] Note: Applicable only for two years. 7. D =[ (P*R2)(300+R)/1003 ] Note:Applicable only for 3 years. 8. [ C.I/(200+R) = S.I/200 ] Note: Applicable only for 2 years. 9. R = [ (2*difference of C.I and S.I)/S.I ] * 100 10.R% amounts after 2 successive years we given:- R = [ (An+1-An)/An ] * 100 An+1 -----> Amount after (n+1) years. An -----> Amount after (n+1) years. 11. P =[ A32/A6 ] = [ A22/A4 ] = [ A12/A2 ] = [A42/A8 ] Note: Double the years. 12. P =[ A23/A32 ] = [ A34/A43 ] = [ A45/A54 ] Note: Consecutive years. P =[ ÖA23/A6 ] = [ ÖA13/A3 ] = [ ÖA33/A9 ] 13. R =[( A6/A3)1/3 - 1 ] = [ ( A4/A2 )1/2 - 1 ] = [ ( A5/A2 )1/3 - 1 ] R =[( A7/A2)1/3 - 1 ] = [ ( A10/A2 )1/8 - 1 ] = [ ( A10/A7 )1/3 - 1 ] 14.Installment problems: a [ 100/(100+R) + 100/(100+R)2 + 100/(100+r)3 + ....... ] = B a -----> Annual installment B -----> Borrowed amount. 15. R = [ (A/P)1/T - 1 ] * 100 16. P = [ A2 * [100/(100+R)]2 ]