yes, a=1,d=2 but there are 50000 terms, so
S = 50000/2 (2+49999(2))
or, since you know the first and last term,
S = 50000/2 (1+99999)
there are 50,000 odd numbers
The sum of the first n odd numbers is n^2.
Would a=1 because its the first odd number in the sequence? And would d=2 ?
So then would you do S99,999= 49999.5(2+99,998(2))
Then divide by two?
S = 50000/2 (2+49999(2))
or, since you know the first and last term,
S = 50000/2 (1+99999)
there are 50,000 odd numbers
The sum of the first n odd numbers is n^2.
In this case, the first odd number in the sequence is 1, and the common difference between consecutive odd numbers is 2. Therefore, yes, a = 1 and d = 2.
To find the sum, we can use the formula for the sum of an arithmetic series:
Sn = (n/2)(2a + (n-1)d)
Here,
- Sn is the sum of the first n terms of the sequence.
- n is the number of terms in the sequence.
- a is the first term.
- d is the common difference.
Applying this formula to our case, we have:
- n = (99,999 - 1) / 2 + 1 = 50,000 (since we are considering odd numbers only)
- a = 1
- d = 2
Substituting these values into the formula:
Sn = (50,000/2)(2*1 + (50,000-1)*2)
= 25,000(2 + 99,998*2)
= 25,000(2 + 199,996)
= 25,000(199,998)
Hence, the sum of all odd numbers up to and including 99,999 is:
Sn = 25,000(199,998) = 4,999,500,000.