find the sum of the integers between 30 and 300 which leaves a reminder 3,when divided by 7?

Asked by riyaliz | 22nd Jul, 2008, 10:47: PM

Expert Answer:

 first  divide 30  by 7 = 4 and remainder 2.    therefore 31 is the first term for the progression  { we want the R=3}

 to find the last term divide 300  by 7 , we will get 42 and R = 6  .  therefore 297   which is the last term  { when divided by 7  leaves 3 as R }

 now we know the first term  and the last term  as 31  and 297  with common difference 7

 the sequence will be 31,38,45,..... 297 

 to calculate  the no . of terms ...  the last term   Tn  = 31 + ( n - 1 ) . 7 = 297

                                                                                                       ( n-1) . 7 = 297 - 31 = 266

                                                                                                        ( n - 1 )   266 / 7 = 38====> n = 38 +1 ==39

 the sum of the integers  ===  31 + 38 + 45 +       ...............297(( 39 terms))

                                S 39 =  39 / 2 ( 31  + 297 )  ==> 39 / 2 ( 328 ) = 39 x 164 =6396

 

  sum of integers between 30 and 300    when divided by 7 leaves 3 as R == 6396

Answered by  | 23rd Jul, 2008, 02:45: PM

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