The sum of the first 12 items of an arithmetic sequence is 354, and the ratio of the sum of the even items to the sum of the odd items in the first 12 items is 32:27. The tolerance D is calculated

The sum of the first 12 items of an arithmetic sequence is 354, and the ratio of the sum of the even items to the sum of the odd items in the first 12 items is 32:27. The tolerance D is calculated

If the first term is A1 and the tolerance is D, then s odd + s even = 354s odd s even = 2732, s even = 192s odd = 162 and s even - s odd = 6D, ∧ d = 5

The sum of the first 12 terms of the arithmetic sequence {an} is 354, the ratio of the sum of odd and even numbers in the first 12 terms is 27:32, and the spherical tolerance is d

d=5
Because it is an arithmetic sequence, odd and even numbers are arranged alternately, that is, the first 12 items have 6 odd and 6 even numbers. Let the sum of odd and even numbers be 27x and 32x, then S12 = 59x = 354, x = 6. The difference between two adjacent odd and even numbers is D, and the difference between odd and even numbers of the first 12 items is 6D = (32-27) x = 5x = 30, then d = 5

Given that there are 20 items in an arithmetic sequence, the sum of odd items is 15, and the sum of even items is 30, then the tolerance is 1

{an}
1 formula a1 + a3 + A5 +... + A19 = 15
2 Formula A2 + A4 + A6 +... + A20 = 30
Type 2-1
30-15=10d
15=10d
Tolerance d = 1.5, that is not 1

It is known that there are 10 items in an equal ratio sequence. If the sum of odd items is 15 and the sum of even items is 30, then the common ratio is ()
A. 5B. 4C. 3D. 2

∵ s odd = a1 + a3 + +A9, s-even = A2 + A4 + +A10 = QS odd, the sum of odd terms is 15, the sum of even terms is 30, q = 2, so D

It is known that a certain arithmetic sequence has 10 times in total, and the sum of odd items is 15 and even items is 30. What is the tolerance D?

Tolerance that the sum of even items minus the sum of odd items is 5 times
(a2-a1)+(a4-a3)+(a6-a5)+（a8-a7）+（a10-a9）
=(a2+a4+a6+a8+a10)-(a1+a3+a5+a7+a9)
=5d
That is 30-15 = 15 = 5D
d=3

The sum of the first 12 items of an arithmetic sequence is 354, and the ratio of the sum of the even items to the sum of the odd items in the first 12 items is 32:27

S odd = [6 (a1 + a11)] / 2 = 6A6
S-even = [6 (a1 + A12)] / 2 = 6a7
Then:
a6/a7=27/32
Also:
Sn=[12(a1＋a12)]/2=6(a6＋a7)=354
The solution is: A6 = 27, a7 = 32, d = a7-a6 = 5

In the arithmetic sequence {an}, given A5 = 10, A12 = 31, find the first term A1 and tolerance D

The tolerance is A1 = * 2-4a1 = 10-4d = 5

In the arithmetic sequence {an}, given A5 = 10, A12 = 31, find the first term A1 and tolerance D of the first term of the sequence {an}

From the meaning of the question, we can get A5 = a1 + 4D = 10, A12 = a1 + 11d = 31, and the solution can get A1 = - 2, d = 3, so the first term of the first term of the sequence {an} is A1 = - 2, and the tolerance is D = 3

In the arithmetic sequence {an}, given a 5 = 10, a 12 ＞ 31, find the range of tolerance D

From the meaning of the title, we can get A12 = (a5-4d) 2 = (10-4d) 2 > 31, we can get - 31 < 4d-10 < 31, we can get the solution of 10 − 314 < d < 10 + 314, and the value range of tolerance D is (10 − 314, 10 + 314)

In known arithmetic sequence {an}, given A5 = 10, A12 = 31, find A1, D, S10

a12=a1+11d=31,a5=a1+4d=10,
So a12-a5 = 7d = 31-10 = 21, so d = 3,
So a1 + 11 × 3 = a1 + 33 = 31, A1 = - 2
So an = - 2 + 3 (n-1) = 3n-5
Sum formula of arithmetic sequence: SN = (a1 + an) n / 2
So S10 = (- 2 + A10) × 10 / 2 = 5 (- 2 + 3 × 10-5) = 5 (- 2 + 25) = 5 × 23 = 115