find the sum of all 3-digit numbers,which leave the remainder 1,when divided by four ?

Asked by riyaliz | 22nd Jul, 2008, 10:03: PM

Expert Answer:

 3 digit  numbers = 100  , 101 ..... 999

    3 digit numbers  which divided by 4 leaves 1 as remainder are   101, 105, 109,  ....................997

   the total  number of terms = = Tn =  101 +  { n-1} 4 == 997

                                                                   101 +4n - 4 = 997

                                                                      97 +4n == 997

                                                                                4n = 900

                                                                             ===> 900 / 4 = 225

   the sum of  225 terms = Sn = 225/ 2  { 2 (101) +( 225-1) 4 ]

                                                       225 [ / 2 [  202+896] ==> 123525

Answered by  | 24th Jul, 2008, 08:14: AM

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