Find the middle term of the sequence formed byall three digit numbers

1.	Find the middle term of the sequence formed byall three digit numbers which leave a remainder 3when divided by 4. Also find the sum of all thenumbers on both sides of the middle term
Answer» The list of 3 digit number that leaves a remainder of 3 when divided by 4 is :103 , 107 , 111 , 115 , .... 999The above list is in AP with first term, a = 103 and common difference, d = 4Let n be the number of terms in the AP.Now, an= 999103 + ( n - 1 ) 4 = 999103 + 4n - 4 = 9994n + 99 = 9994n = 900n = 225 Since, the number of terms is odd, so there will be only one middle term.middleterm=(n+12)thterm=113thterm=a+112d=103+112×4=551 Weknowthat,sumoffirstntermsofanAPis,Sn=n2[2a+(n−1)d] Now,Sum=112/2[2×103+111×4]=36400 Sumofalltermsbefore middleterm=36400sum of all numbers= 225/2[2×103+224×4]=123975 Now,sumoftermsafter middleterm=S225−(S112+551)=123975−(36400+551)=87024

Find the middle term of the sequence formed byall three digit numbers which leave a remainder 3when divided by 4. Also find the sum of all thenumbers on both sides of the middle term

Answer»

The list of 3 digit number that leaves a remainder of 3 when divided by 4 is :103 , 107 , 111 , 115 , .... 999The above list is in AP with first term, a = 103 and common difference, d = 4Let n be the number of terms in the AP.Now, an= 999103 + ( n - 1 ) 4 = 999103 + 4n - 4 = 9994n + 99 = 9994n = 900n = 225

Since, the number of terms is odd, so there will be only one middle term.middleterm=(n+12)thterm=113thterm=a+112d=103+112×4=551 Weknowthat,sumoffirstntermsofanAPis,Sn=n2[2a+(n−1)d] Now,Sum=112/2[2×103+111×4]=36400 Sumofalltermsbefore middleterm=36400sum of all numbers= 225/2[2×103+224×4]=123975

Now,sumoftermsafter middleterm=S225−(S112+551)=123975−(36400+551)=87024

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