The first term of an arithmetic sequence is 6, and the tenth term is 2.
(a) Find the common difference d.
d =
Find the 100th term of the sequence.
a100 =
(b) Find the partial sum of the first ten terms.
S10 =
how do i do this, there's an example I tried following but it did not ask me for the 100th term nor the first ten terms, please ayuda?
Oh, math problems, my favorite! Let's solve this in a fun way, shall we?
(a) To find the common difference (d), we can use the formula a_n = a_1 + (n - 1) * d, where a_n represents the nth term. In this case, the first term (a_1) is 6, and the tenth term (a_10) is 2. Plug those values into the formula: 2 = 6 + (10 - 1) * d. Now let's solve for d:
2 = 6 + 9d
-4 = 9d
d = -4/9
So, the common difference (d) is -4/9. Don't worry, math can be negative sometimes, but it's still good at heart!
(b) Now, let's find the 100th term (a100). We'll use the same formula as before:
a_n = a_1 + (n - 1) * d
a100 = 6 + (100 - 1) * (-4/9)
Calculating the above equation gives us:
a100 = -346/9
Therefore, the 100th term is -346/9. Math can be a bit of a joker sometimes!
Now, let's tackle the partial sum of the first ten terms (S10). We can use the formula for the sum of an arithmetic series:
S_n = (n/2) * (2a_1 + (n - 1) * d)
Plugging in the values:
S10 = (10/2) * (2 * 6 + (10 - 1) * (-4/9))
Simplifying the above equation gives us:
S10 = 30
So, the partial sum of the first ten terms (S10) is 30. Math can also have a sweet side, right?
I hope that clears things up and brings a smile to your face! Remember, math problems are just puzzles waiting to be solved. Feel free to ask if you need any more help!
To find the common difference (d) in an arithmetic sequence, you can use the formula:
d = (10th term - 1st term) / (n - 1)
Here, the first term is 6 and the tenth term is 2.
(a) Let's calculate the common difference:
d = (2 - 6) / (10 - 1)
d = -4 / 9
d = -4/9
So, d = -4/9.
To find the 100th term (a100) of the sequence, you can use the formula:
a100 = 1st term + (n - 1) * d
Here, the first term is 6 and the common difference is -4/9.
(a100) = 6 + (100 - 1) * (-4/9)
Let's calculate the value:
a100 = 6 + 99 * (-4/9)
a100 = 6 - 44
a100 = -38
So, a100 = -38.
(b) To find the partial sum of the first ten terms (S10), you can use the formula:
S10 = (n/2) * (2a + (n - 1) * d)
Where n is the total number of terms.
Here, n = 10, a = 6, and d = -4/9.
Let's calculate the partial sum:
S10 = (10/2) * (2 * 6 + (10 - 1) * (-4/9))
S10 = 5 * (12 - (9) * (4/9))
S10 = 5 * (12 - 4)
S10 = 5 * 8
S10 = 40
So, the partial sum of the first ten terms (S10) is 40.
To find the common difference (d) in an arithmetic sequence, you can use the formula:
d = (term_n - term_1) / (n - 1)
where term_n is the nth term of the sequence, term_1 is the first term, and n is the position of the term.
Given that the first term (term_1) is 6 and the tenth term (term_10) is 2, you can use these values to find the common difference (d):
d = (2 - 6) / (10 - 1)
Simplifying this expression, you have:
d = -4 / 9
So, the common difference is -4 / 9.
To find the 100th term (term_100) of the sequence, you can use the formula for the nth term in an arithmetic sequence:
term_n = term_1 + (n - 1) * d
Given that term_1 is 6, d is -4 / 9, and n is 100, you can substitute these values into the formula:
term_100 = 6 + (100 - 1) * (-4 / 9)
Simplifying this expression, you have:
term_100 = 6 + 99 * (-4 / 9)
term_100 = 6 - 44
term_100 = -38
So, the 100th term of the sequence is -38.
To find the partial sum of the first ten terms (S10), you can use the formula:
S10 = (n / 2) * (term_1 + term_n)
Given that n is 10, term_1 is 6, and term_n is 2, you can substitute these values into the formula:
S10 = (10 / 2) * (6 + 2)
S10 = 5 * 8
S10 = 40
So, the partial sum of the first ten terms is 40.
a(n) = a(1) + (n-1) d
a(10) = 2 = 6 + 9 d
-4 = 9 d
d = -4/9