If S10 = 4s5, then A1: D equals () A. 14B. 12C. 2D. 4

If S10 = 4s5, then A1: D equals () A. 14B. 12C. 2D. 4

Let A1 = a, S10 = 10A + 10 (10 − 1) D2 = 10A + 45d, S5 = 5A + 5 (5 − 1) D2 = 5A + 10d, from S10 = 4s5, 10a + 45d = 4 (5a + 10d), i.e., 10a + 45d = 20A + 40d, ∵ 5D = 10a, ∵ D ≠ 0, ∵ ad = 510 = 12, i.e., A1: D = 12

In the arithmetic sequence {an}, A1 = 10, S10 = 5s5, then the tolerance d =?
The whole process Thank you

s10=(a1+a10)*10/2=(a1+a1+9d)*5=(2a1+9d)*5=(20+9d)*5=100+45ds5=(a1+a5)*5/2=(a1+a1+4d)*5/2=(a1+2d)*5=(10+2d)*5=50+10dS10=5S5100+45d=5(50+10d)100+45d=250+50d5d=-150d=-30

Let the sum of the first n terms of the arithmetic sequence {an} be SN. If S10: S5 = 1:2, then the value of S15: S5 is

S5, s10-s5, s15-s10 are still equal
Because S10: S5 = 1:2
Let S10 = x, then: S5 = 2x
At the same time, let S15 = y
Then: 2x, x-2x, Y-X are equal difference
That is: 2x, - x, Y-X are equal difference
Yide: Y-X = - 4x
So, y = - 3x
So, S15: S5 = - 3 / 2

Let A1 and d be real numbers, the first term be A1, the sum of the first n terms of the arithmetic sequence {an} with tolerance d be Sn, satisfying s5s6 + 15 = 0, then the value range of D is______ ．

Because s5s6 + 15 = 0, so (5A1 + 10d) (6A1 + 15d) + 15 = 0, we can get 2A12 + 9a1d + 10d2 + 1 = 0. This equation can be regarded as a quadratic equation with one variable about A1, it must have roots, so △ = (9D) 2-4 × 2 × (10d2 + 1) = d2-8 ≥ 0, we can get D2 ≥ 8, we can get D ≥ 22, or D ≤ - 22, then D is taken as

Let the sum of the first n terms of the arithmetic sequence {an} be Sn, and S5 = - 5, S10 = 15, find the sum of the first n terms of the sequence {Sn / N} and TN

S5 = 5A3, so A3 = - 1
S10-s5 = A6 +... + A10 = a1 +... + A5 + 5 times 5D, d = 1
So A1 = minus 3
an=n-4
Sn=0.5n^2-3.5n
Sn/n=0.5n-3.5
Tn=n（n-13）/4

Arithmetic sequence, tolerance d = 3, A5 = 11, calculate sequence A10 and S10
Detailed steps and standard formula are required

a5=a1+4d=a1+4×3=11
So A1 = - 1
a10=a1+9d=-1+9×3=26
s10=（-1+26）×10×0.5=125

In known arithmetic sequence an, A10 = 10, S10 = 70, d =?

s10=(a1+a10)*10/2 a1=4
d=（a10-a1)/9=2/3

Given the tolerance d = 2, the first 10 terms and S10 = 40, then A2 + A4 + A6 + A8 + A10 =?

25

Let the arithmetic sequence consist of three numbers, the sum of three items is 21, and the sum of squares of each item is 179

Let these three numbers be a, B, C, because the three numbers form an arithmetic sequence, so 2B = a + C and a + B + C = 21, B + 2B = 21, B = 7, so C = 14-A because a & sup2; + B & sup2; + C & sup2; = 179, that is, a & sup2; + 7 & sup2; + (14-A) & sup2; = 179a & sup2; + 49 + 196-28a + A & sup2; = 1792a & sup2; - 28a + 66 = 0A & sup2; - 14

Tolerance problem of arithmetic sequence
The sum of the first 12 terms of the arithmetic sequence {an} is 354, and the ratio of the sum of the odd and even terms in the first 12 terms is 27:32
The idea in the answer: the odd and even items in the first 12 items constitute a new arithmetic sequence with A1 and A2 as the first items and 2D as the tolerance. S odd = 6A1 + (6 * 5) / 2 * 2D = 6A1 + 30d
If you don't understand why the tolerance is 2D, explain the reason

a(n+2)-an=[a1+(n+2-1)d]-[a1+(n-1)d]=2d