It is known that a, B, X1 and X2 are mutually unequal positive numbers, Y1 = (ax1 + bx2) / (a + b), Y2 (BX1 + AX2) / (a + b), try to analyze the size of y1y2 and x1x2, and say

It is known that a, B, X1 and X2 are mutually unequal positive numbers, Y1 = (ax1 + bx2) / (a + b), Y2 (BX1 + AX2) / (a + b), try to analyze the size of y1y2 and x1x2, and say

YIY2>X1X2

The first term of the positive term sequence {an}, A1 = 1 / 2, the sum of the first n terms is Sn, (the 10th power of 2 multiplied by S30) - (the 10th power of 2 + 1) S20 + S10 = 0 to find the sum of the first n terms of {NSN}

2^10*（S30-S20）=S20-S10
2^10*a21（1-q^10)/(1-q)=a11(1-q^10)/(1-q)
2^10*a21=a11
q=1/2
an=（1/2）*（1/2）^（n-1）=（1/2）^n
Sn=1-(1/2)^n
nSn=n-n*(1/2)^n
Tn=n*(n+1)/2+(1/2)^(n+1)-2

The first term of the positive proportional sequence {an}, A1 = 1 / 2, the sum of the first n terms is Sn, (the 10th power of 2 multiplied by S30) - (the 10th power of 2 + 1) S20 + S10 = 0
Finding the general term formula of sequence an finding the sum TN of the first n terms of sequence n * SN

2^10*（S30-S20）=S20-S102^10*a21（1-q^10)/(1-q)=a11(1-q^10)/(1-q)2^10*a21=a11q=1/2an=（1/2）*（1/2）^（n-1）=（1/2）^nSn=1-(1/2)^nnSn=n-n*(1/2)^nTn=n*(n+1)/2+(1/2)^(n+1)-2

If S30 = 13s10, S10 + S30 = 140, then the value of S20 is______ ．

Because S30 = 13s10, S10 + S30 = 140, so S10 = 10, S30 = 130. ∵ sequence {an} is an equal ratio sequence, ∵ Sn, s2n Sn, s3n-s2n are also equal ratio sequences, that is, S10, s20-s10, s30-s20 are also equal ratio sequences, so S20 = 40, or S20 = - 30, because S20 = S10 (1 + Q10), so S20 = 40

Let Sn be the sum of the first n terms of the equal ratio sequence, if S10 = 10, S20 = 30, then S30=______ ．

According to the properties of the equal ratio sequence, S10, s20-s10, s30-s20 are equal ratio sequence ｛ (s20-s10) 2 = S10 · (s30-s20) ｛ 400 = 10 (s30-30) ｛ S30 = 70, so the answer is: 70

Let Sn be the sum of the first n terms of the equal ratio sequence, if S10 = 10, S20 = 30, then S30=______ ．

According to the properties of the equal ratio sequence, S10, s20-s10, s30-s20 are equal ratio sequence ｛ (s20-s10) 2 = S10 · (s30-s20) ｛ 400 = 10 (s30-30) ｛ S30 = 70, so the answer is: 70

Let Sn be the sum of the first n terms of the equal ratio sequence, if S10 = 10, S20 = 30, then S30=______ ．

According to the properties of the equal ratio sequence, S10, s20-s10, s30-s20 are equal ratio sequence ｛ (s20-s10) 2 = S10 · (s30-s20) ｛ 400 = 10 (s30-30) ｛ S30 = 70, so the answer is: 70

Let Sn be the sum of the first n terms of the equal ratio sequence, if S10 = 10, S20 = 30, then S30=______ ．

According to the properties of the equal ratio sequence, S10, s20-s10, s30-s20 are equal ratio sequence ｛ (s20-s10) 2 = S10 · (s30-s20) ｛ 400 = 10 (s30-30) ｛ S30 = 70, so the answer is: 70

In the arithmetic sequence {an}, A1 = - 2012, and the sum of the first n terms is SN. If a12-a10 = 2, then the value of s2012 is equal to - 2012

S12 / 12 is the average of the first 12 items in the sequence, namely (a1 + A12) / 2
Similarly, S10 / 10 = (a1 + A10) / 2
The two formulas are subtracted, that is, (a12-a10) / 2 = 2, so d = 2
So s2012 = 2012a1 + 2012 × (2012-1) × 2 / 2 = 2012 × (- 2012) + 2012 × 2011 = - 2012

In the arithmetic sequence an, if S10 = 20, AN-9 + an-8 +. + an = 50, Sn = 56, then the value of n is
N must be less than 20, then what's the idea?

S10 = a1 + A2 +. + A10 = 20 ① AN-9 + an-8 +. + an = 50 write it backwards: an + a (n-1) +. + a (N-9) = 50 ② (this is the sum of the last 10 items) note that a1 + an = A2 + a (n-1) =. = A10 + a (N-9) ① + ② get: 10 (a1 + an) = 70 ∵ a1 + an = 7 ∵ Sn = (a1 + an) * n / 2 = 56 ∵ 7; n / 2 = 56 ∵ n = 16