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Find the middle term of the sequence formed by all three-digit numbers which leavea remainder 3. When divided by 4. Also find the sum of all numbers on both sides ofthe middle term separately.24. |
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Answer» The list of 3 digit number that leaves a remainder of 3 when divided by 4 is : 103 , 107 , 111 , 115 , .... 999 The above list is in AP with first term, a = 103 and common difference, d = 4 Let n be the number of an the AP. Now, an= 999 103 + ( n - 1 ) 4 = 999 103 + 4n - 4 = 999 4n + 99 = 999 4n = 900 n = 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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