Find the middle term of the sequence formed by all the numbers between 9and 95which leaves a remainder 1 when divided by 3.Also find the sum of numbers on both the sides of the middle term separately
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Numbers leaving a remainder of 1 when divided by 3 can be expressed as:-
3*m + 1
Smallest such number in the given range is 10 = 3*3 +1 = 7 + 3*1
Largest such number in the given range is 94 = 3*31 +1 = 7 + 3*29
So as you can see there are 29 terms
The middle term would be the 15th term. = 7 + 3*15 = 52.
Hence 52 is the first answer.
So on the left side start and end terms are 10......49
The Nth term is 7 + 3*n
Hence the sum of first 14 terms is 7*14 + 3*14*15/2 = 413
The sum of all 29 terms is 7*29 + 3*29*30/2 = 1508
Hence the sum of the last 14 terms is = 1508 - 413 - 52 = 1043
Hence the answers are
52
413
1043
3*m + 1
Smallest such number in the given range is 10 = 3*3 +1 = 7 + 3*1
Largest such number in the given range is 94 = 3*31 +1 = 7 + 3*29
So as you can see there are 29 terms
The middle term would be the 15th term. = 7 + 3*15 = 52.
Hence 52 is the first answer.
So on the left side start and end terms are 10......49
The Nth term is 7 + 3*n
Hence the sum of first 14 terms is 7*14 + 3*14*15/2 = 413
The sum of all 29 terms is 7*29 + 3*29*30/2 = 1508
Hence the sum of the last 14 terms is = 1508 - 413 - 52 = 1043
Hence the answers are
52
413
1043
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