Given the arithmetic sequence {an}, Sn is the sum of the first n terms, and S10 = S20, then S30=______ ．

Given the arithmetic sequence {an}, Sn is the sum of the first n terms, and S10 = S20, then S30=______ ．

∵ arithmetic sequence {an}, S10 = S20, let S10 = S20 = a, S30 = B, ∵ a, 0, B-A be arithmetic sequence, ∵ 0 = a + B-A, B = 0

If SN is the sum of the first n terms of the arithmetic sequence {an}, and S10 = 5, S20 = 17, then S30 =?, S20 =

S10=5,S20=17
S20-S10=12
Because SN is the sum of the first n terms of the arithmetic sequence {an}
So S10 s20-s10 s30-s20 It is also an arithmetic sequence
S30-S20=19
S30=19+17=36

Let the sum of the first n terms of the arithmetic sequence {an} be Sn, and A3 = 24 and Sn = 0 are known, then the general term formula of the sequence {an} can be obtained

There is a formula for the arithmetic sequence. Just bring it in according to the formula. The simultaneous equation Sn = (a1 + an) / 2 * n an = A3 + (n-3) d

Let the sum of the first n terms of the arithmetic sequence {an} be SN. If S10 = 30, S20 = 100, find S30=___________

If an is an arithmetic sequence, then Sn, s2n Sn, s3n-s2n. Are also arithmetic sequences
So 30100-30, s30-100 is an arithmetic sequence
（s30-100）-70=40
s30=210

Function f (x) = {5 ^ | X-1 | - 1 (x > 0) x ^ 2 + 4x + 4 (x > 0) with domain R

Sketch the function f (x)
x

What is the domain of the function f (x) = 1-4x?

The domain of the function f (x) = 1-4x is all real numbers

The interval where y = TaNx and y = SiNx are monotone increasing functions at the same time is

[- π + 2K π, π / 2 + 2K π] (k is an integer)

Let x belong to R, the inequality x ^ 2log2 be the bottom 4 (a + 1) / A + 2xlog2 be the bottom 2A / A + 1 + log2 be the bottom (a + 1) ^ 2 / 4A ^ 2 > 0 hold, and find the value range of A

Let ㏒ 2 ((a + 1) / a) = t, then ㏒ 2 [4 (a + 1) / a] = 2 + T ㏒ 2 (2A / (a + 1)) = 1-T ㏒ 2 [(a + 1) & # 178; (4a & # 178;] = 2 (t-1). It can be changed into: (2 + T) x & # 178; + 2 (1-T) x + 2 (t-1) > 0, so it only needs 2 + T > 0, and △ 1 is ㏒ 2 ((a + 1) / a > 1 (a + 1) / a > 2 ㏒ 0

The solution of the inequality log2 (x + 1) > 1 / 2log2 [a (X-2) + 1] (a > 2)
2 and 4 are base numbers

If the domain x + 1 > 0, a (X-2) + 1 > 0, then x ＞ - 1, X ＞ 2-1 / A, so x ＞ 2-1 / A
log2（x+1）>1/2log2[a(x-2)+1]
2log2（x+1）>log2[a(x-2)+1]
log2（x+1）²>log2[a(x-2)+1]
（x+1）²>a(x-2)+1
x²+2x+1>ax-2a+1
x²+（2-a）x-2a＞0
（x+2a）(x-a)＞0
Because a > 2, x > A or x < - 2A
In conclusion: x > a

Solving log2 (a ^ 2 + 1) with logarithmic inequality
Correct, the question should be log2 (a ^ 2 + 1)

a^2+1