Use the arithmetic sequence of numbers 1, 3, 5, 7, 9,…to find the following:?

Question

What is d, the difference between any 2 terms
Using the formula for the nth term of an arithmetic sequence, what is 101st term
Using the formula for the sum of an arithmetic series, what is the sum of the first 20 terms?

Answer 1

the formula is

y = 2(n - 1) + 1
y = 2n - 2 + 1
y = 2n - 1

the difference between them is 2.

y = 2(101) - 1
y = 202 - 1
y = 201

ANS : 201

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Sn = n/2 (a1 + an)
S(20) = (20/2)(1 + 2(20) - 1)
S(20) = 10(40)
S(20) = 400

ANS : 400

Answer 2

The pattern for this sequence is (2n-1) where n is the term. The first term is [2(1)-1] = 1, the second term is [2(2)-1] = 3, the third term is [2(3)-1] = 5, and so forth.

The difference between any two consecutive terms is obviously 2. I assume, however, that what you are seeking is an equation to calculate (for example) the difference between the eighth and third terms. Let m be the first of the terms (the third term in my exampe) and n be the second (the eighth term in my example). The difference would be
(2n - 1) - (2m - 1) = 2n - 2m = 2(n - m)

The 101st term would be [2(101)-1] = 201

The sum of the first twenty terms can be calculated as well. First, note that the twentieth term is [2(20)-1] = 39. So
20(39 + 1)/2 = 400
This uses the formula [n(a1+an)]/2 where a1 is the first term and an is the nth term.

Answer 3

This is a very basic plug and chug question:
The formula for the nth term of an arithmetic seq. is
An= A0+d(n-1)
Where d is the difference between two consec. terms
A0 is the first term
And An is the nth term

So.. An = 1+2(101-1)
An=201

The sum formula is:
Sn=(n/2)(A0+An)
(20/2)(1+[1+2(20-1)])
(10) (40) = 400

Answer 4

The quick way to sum an arithmetic sequence is to add up two copies of it and then divide by two. It's easier than adding up one copy.

The 20th term of this sequence is (20*2)-1 or 39. If we write 2 copies of the sequence, one of them backwards, we get:

1, 3, 5, 7, .... 37, 39
39, 37, 35, 33, .... 3, 1

If you notice, each number added to the one below it is 40. And since there are 20 terms, we get 20 40's, which adds up to 800.

But 800 is the sum of TWO copies of the sequence, so we divide that by two to get the sum of ONE copy of the sequence.

Answer 5

d = a_n - a_(n - 1)
d = 3 - 1 = 5 - 3 = 7 - 5 = 9 - 7
d = 2

The formula for nth term (a_n) is
a_n = a_1 + (n - 1)d
a_n = 1 + (n - 1)(2)
a_n = 1 + 2n - 2
a_n = 2n - 1

101st term:
a_101 = 2(101) - 1
a_101 = 202 - 1
a_101 = 201

Sum of arithmetic series
s_n = n(a_1 + a_n)/2
s_n = n[1 + (2n - 1)]/2
s_n = n(1 + 2n - 1)/2
s_n = n(2n)/2
s_n = n(n)
s_n = n²

Sum of first 20 terms:
s_20 = 20²
s_20 = 400

^_^

Answer 6

d=2

an=dn+1-d
=2n-1
a101=201

sigman=(an+a1)xn/2
=(2n-1+2x1-1)xn/2
=n^2
sigma20=400

Answer 7

d = 2, for any CONSECUTIVE terms
t(101) = (101 * 2) - 1 = 201
uhh... i dunno, but the answer is 20², or 400.

Answer 8

wat do u mean i dont understand