In the arithmetic sequence an, if the sum of the first n terms is Sn, S10 = 20, S20 = 30, then S30=

In the arithmetic sequence an, if the sum of the first n terms is Sn, S10 = 20, S20 = 30, then S30=

Arithmetic sequence
S10 s20-s10 s30-s20 is also an arithmetic sequence
20 10 0
therefore
S30=S20=30

SN is the sum of the first n terms of the arithmetic sequence {an}, given S10 = 100, S20 = 10, find S30

Substituting S10 = 100 and s20-10 into the equation Sn = Na1 + [n (n-1) D] / 2
Can get
10A1 + [10 (10-1) D] / 2 = 100. Formula 1
20a1 + [20 (20-1) D] / 2 = 10. Formula 2
The first figure is A1 = 18.55
The tolerance is d = - 1.9
Then S30 = 30a1 + [30 (30-1) D] / 2 = 30 * 18.55 + (29 * 30 * D / 2) = - 270

On the inequality 2 | log of X with 2 as the base, the logarithm | + a ≥ 0 of X always holds for any x ∈ (0, + ∞), and the value range of a is obtained
Answer process````````

In addition, y = | log takes the logarithm of 2 as the base x | to make an image, and it is found that the logarithm of 2 as the base x | of | log is always greater than or equal to 0, this number multiplied by 2, you don't care about it, the coefficient doesn't care about it, and then you can achieve 2Y + a greater than or equal to 0 only when a is greater than or equal to 0 from the image translation

If log2 (- x) < x + 1 holds, the value range of X is______ ．

It can be judged by drawing that f (x) = log2 (- x) and G (x) = x + 1 intersect at (- 1,0). The former is monotonically decreasing, and the latter is monotonically increasing. Therefore, log2 (- x) < x + 1 holds only when - 1 < x < 0, so the answer is: (- 1,0)

Given the function f (x) = x2-6x + 8, X ∈ [1, a], and the minimum value of function f (x) is f (a), then the value range of real number a is______ ．

Function f (x) = x2-6x + 8 = (x-3) 2-1, X ∈ [1, a], and the minimum value of function f (x) is f (a), and ∵ function f (x) monotonically decreases in the interval [1, 3], so the answer is: (1, 3]

If y = (MX ^ 2 + 4 √ 3 + n) / (x ^ 2 + 1), (M

y=(mx^2+4√3x+n)/(x^2+1)
(m-y)x^2+4√3x+n-y=0
The discriminant △≥ 0 for the unknown number x of the upper equation
That is, (4 √ 3) ^ 2-4 (M-Y) * (n-y) ≥ 0
y^2-(m+n)y+mn-12≤0
Because the value range of the function is [- 1,7],
That is to say, the solution set of the above inequality is [- 1,7],
Because the end point of the solution set of quadratic inequality with one variable is the root of the corresponding equation,
So - 1 and 7 are two parts of the equation y ^ 2 - (M + n) y + Mn-12 = 0,
∴-1+7=m+n,
-1*7= mn-12,
The simultaneous solution is m = 1, n = 5

A machine has done a total of 8000j of work, of which the extra work is 13% of the active work, so what is the active work of this machine___ What is the mechanical efficiency___ ．

W you + W amount = w total, w you + 13W you = 8000j, so w you = 6000J, η = w you, w total = 6000J, 8000j = 75%

Ask three practical questions (Grade 5)
(1) There are 12 soldiers on guard in a border station. Three soldiers are sent to stand guard in turn every hour, 24 hours a day. How many hours does each soldier stand on guard?
(2) Mother bought 3kg bananas and 5kg apples, and paid 41.20 yuan in total. Every kilogram of bananas is 1.2 yuan cheaper than every kilogram of apples. How much is it per kilogram of bananas?
(3) A shopping mall sells 700 bags of rice in the morning and 300 bags of the same rice in the afternoon, 4000 kg more in the morning than in the afternoon?

There are 12 soldiers on guard in a border station. Three soldiers are sent to stand guard in turn every hour, 24 hours a day. How many hours does each soldier stand on guard?
The total time of standing guard is 3 * 24 = 72 hours
The guard time of each soldier is 72 / 12 = 6 hours
2) Mother bought 3kg bananas and 5kg apples, and paid 41.20 yuan in total. Every kilogram of bananas is 1.2 yuan cheaper than every kilogram of apples. How much is it per kilogram of bananas?
3 kg bananas and 5 kg apples, all as bananas, 5 * 1.2 more = 6 yuan
So (41.2-6) / (3 + 5) = 4.4 yuan per kilogram of banana
A shopping mall sells 700 bags of rice in the morning and 300 bags of the same rice in the afternoon, 4000 kg more in the morning than in the afternoon?
Solution: x kg per bag of rice
700x-300x=4000
400x=4000
X = 4000 divided by 400
X = 10 (kg)
Morning: 10x700 = 7000 (kg)
Afternoon: 10x300 = 3000 (kg)
A: 7000 kg in the morning,
Sell 3000 kg in the afternoon

An ammeter with 500 microampere range and 300 ohm internal resistance has a 150 ohm resistance in parallel at both ends of the ammeter. How to calculate how to expand the range of the ammeter to several ma

1.5mA
500uA + 500uA * 300eu / 150eu

The school plans to buy 40 pens and some notebooks (the number of notebooks exceeds the number of pens). The price of both stores is 10 yuan for each pen and 2 yuan for each notebook. The preferential way of store a is 10% off for pens and 8% off for notebooks. The preferential way of store B is to give 1 notebook for every 5 pens, and there is no discount for pens, 50% off for notebook purchase. Within what range is it more cost-effective to go to a store?
Don't just know formula and calculation

Let's buy notebook x (x > 40) to buy 40 pens in a store, X notebook need: 40 * 10 * 0.9 + 2 * x * 0.8 yuan --- this should be clear at a glance, to buy 40 pens in B store, X notebook need: 40 * 10 + 2 (X-8) * 0.75 --- explain: because buy 5 pens to send a book, to buy 40 steel