We're working on geometric and athrimetic series and I really don't know where to start with this except that t sub 1 = 101
How would I find the sum of the multiples of 5 between 101 and 2001 help
well, the first multiple of 5 in that interval is 105.
The last is 2000.
105 + 110 + 115 + ........ 2000
n-1 = (2000-105)/5 = 379
n = 380
Sum = (n/2)(a1+an) = (380/2)(2105)
=399,950
Thank you very much
You are welcome :)
To find the sum of the multiples of 5 between 101 and 2001, we can use the formula for the sum of an arithmetic series. Here are the steps you can follow to solve the problem:
1. Identify the first term and the last term of the series:
- In this case, the first term (t₁) is given as 101.
- The last term (tₙ) is the largest multiple of 5 that is less than or equal to 2001. To find this, divide 2001 by 5 and round down to the nearest whole number. This will give you the number of terms in the series (n). Then, multiply this number by 5 to find the last term.
tₙ = (2001 ÷ 5) × 5 = 400 × 5 = 2000.
2. Calculate the number of terms (n):
- The number of terms can be found by subtracting the first term from the last term and then dividing the result by the common difference (which is 5 in this case).
n = (tₙ - t₁) ÷ d = (2000 - 101) ÷ 5 = 1899 ÷ 5 = 379.
3. Calculate the sum using the formula for the sum of an arithmetic series:
- The formula for the sum of an arithmetic series is:
S = (n/2) × (t₁ + tₙ),
where S is the sum, n is the number of terms, t₁ is the first term, and tₙ is the last term.
- Plug in the values and solve for the sum:
S = (379/2) × (101 + 2000) = 189.5 × 2101 = 398974.5.
Therefore, the sum of the multiples of 5 between 101 and 2001 is 398974.5.