The sum of the integers 1 to 99 is 4950. An easy way to figure this out is using the equation N*(N+1)/2 where N is the largest number in the set.
100
the series of odd numbers from 1 to 99 :1 3 5 7 9.....99 SUM OF THE SERIES: It is a geometric progression with a=1 and l=99 and common difference (d)=2. let 99 be the nth term of the sequence so, 99=1+(n-1)d 99=1+(n-1)2 solving this we get n =50. SUM=(n/2)(a+l) =(50/2)(1+99) =(25)(100) =2500.
9801 * The sum of the first two odd numbers (1+3) is 4, or 22 * The sum of the first three odd numbers (1+3+5) is 9, or 32 * The sum of the first four odd numbers (1+3+5+7) is 16, or 42 * ...and so on So the sum of the first 99 odd numbers, using the pattern above, would be 992 or 9801.
let the number be n-1, n, n+1 then 3n=99 and so n=33 the numbers are 32, 33, 34
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the series of odd numbers from 1 to 99 :1 3 5 7 9.....99 SUM OF THE SERIES: It is a geometric progression with a=1 and l=99 and common difference (d)=2. let 99 be the nth term of the sequence so, 99=1+(n-1)d 99=1+(n-1)2 solving this we get n =50. SUM=(n/2)(a+l) =(50/2)(1+99) =(25)(100) =2500.
9801 * The sum of the first two odd numbers (1+3) is 4, or 22 * The sum of the first three odd numbers (1+3+5) is 9, or 32 * The sum of the first four odd numbers (1+3+5+7) is 16, or 42 * ...and so on So the sum of the first 99 odd numbers, using the pattern above, would be 992 or 9801.
The sum of 1+ 3 + 5 + ... + 99 = 50 x (1 + 99) ÷ 2 = 2500.
let the number be n-1, n, n+1 then 3n=99 and so n=33 the numbers are 32, 33, 34
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2500
Do you mean:"What three consecutive numbers add up to 99?" If you do, then the numbers are 32, 33, 34.
The numbers are 97, and 99.
s100=50(2+99) this is because a=1, d=1 and n=100. answer is 5050