SOLUTION: I've got this problem (see below). I've worked through problems 1b - 1d, but I am stuck with how to answer 1e. Can someone help? Thanks A Bunch! 1) Use the arithmetic sequ

Question 44629: I've got this problem (see below). I've worked through problems 1b - 1d, but I am stuck with how to answer 1e. Can someone help?
Thanks A Bunch!
1) Use the arithmetic sequence of numbers 1, 3, 5, 7, 9,�to find the following:
a) What is d, the difference between any 2 terms?
Answer: d = 2
Show work in this space.
The difference between any two terms = (n+1 ) th term - (n) th term
So, the difference d = 3-1 = 5-3 = 7-5 = 2
Answer d = 2

b) Using the formula for the nth term of an arithmetic sequence, what is 101st term?

Answer: nth term = 201

Show work in this space.
Given a = first term = 1
n = 101
d = 2
putting above values in formula 1
nth term = 1 + (101-1) * 2
= 1 + 200 = 201
So, nth term = 201

c) Using the formula for the sum of an arithmetic series, what is the sum of the first 20 terms?
Answer: 400
Show work in this space
n = 20
a = first term = 1
So, sum = 20 / 2 (2 * 1 + 19 * 2)
= 10 * (40)
sum = 400

d) Using the formula for the sum of an arithmetic series, what is the sum of the first 30 terms? =n/2(a +(n-1)d)��..(1)
Answer: 900
Show work in this space
given that n = 30
a = first term =1
= 30 / 2 (2 * 1 + 29 * 2)
=15 * (60)
= 900

e) What observation can you make about these sums of this series (HINT: It would be beneficial to find a few more sums like the sum of the first 2, then the first 3, etc.)? Express your observations as a general formula in "n."
Answer:

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
When "d" is a positive number the sum of terms gets
increasingly greater as you add more terms.
The sum of "n" terms is n times the average of the
1st and the nth term, i.e. S(n)=(n)[a+(n-1)d]/2
Cheers,
Stan H.

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