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18, trust me since you must combine like terms, then plug in two variables to group terms, and finally multiply the value of the variables after you compare the two expressions that you get by first combining like terms, then grouping terms with variables.

s10  =   10a1  + (10 * 9)d / 2

s20   =   20a1  + (20 * 19)d / 2

So

0  =  10a1  + 45d    →  0  = -20a1 - 90d       (1)

1/5  = 20a1  +190d      (2)

add (1) , (2)

1/5  = 100d

d = 1/500

And

0  = 10a1 + 45(1/500)

10a1  =  -45/500

a1  = -45/5000  =  -9/1000

s70  =  70(-9/1000)  + (70 * 69)/2 * (1/500) =    -630/1000 +  2415 / 500  = 21 / 5 

\(2 + \frac{6}{7} +\frac {18}{49} + \frac{54}{343} + 1 + \frac{2}{14} + \frac{3}{98} \)

Rearrange as

1 + 2 + 6/7 + 2/14 + 18/49 + 54/343 + 3/98   = 

1 + 2 + (6/7 + 1/7)  +  18/49 + 54/343 + 3/98   = 

1 + 2 + 1  +   (126 + 54) / 343  +   (1.5) / 49   =

4  + 180/343  +  7(1.5) / 343   = 

4 + (180 / 343) + (10.5) /343  =

4 + (1800 + 105)  / 3430  

4 + 1905 /3430

4 + 381/686  =

3125 / 686

.036 / .4  = 9/100

,0000324 / .036  = 9 /10000

Not geometric !!!!

BC = 1

AC = sqrt 3

AB = 2

Construct a circle at B = (1,0)  with a radius of sqrt 2

The equation of the circle  is   (x -1)^2 + y^2  = 2

To find the intersection of BD and AC  let  x = 0

Then y  = 1

So D =  (x ,y)  = (0, 1)

BD  =  radius of the circle  = sqrt 2

DC  =1

The probability that BD < sqrt 2      is       DC / AC   =  1 / sqrt 3 ≈  57.7 %

https://web2.0calc.com/questions/algebra_19078

Sum =  [first term + last term ] * [number of terms / 2]

Nuimber of terms =   [6021 - 21 ]  / 6   +  1    =  1001

Sum =   [ 21 + 6021 ] *  [ 1001 / 2 ] =   3,024,021

The sum of the expression is 13681

.

The sum of the series is

 over 49 end-fraction

229/49

.

The value of the expression is

7381+2481i73 over 81 end-fraction plus 24 over 81 end-fraction i

7381+2481𝑖

See my previous answer to your \(a_{12}\) question. It should be a similar (if not the exact same) process with Cauchy Schwarz. 

Edit: my previous answer got large ai language modeled. 

Here is how to start with cauchy: \(x_i = \sqrt{i} \implies \sum_{i=1} ^{20}{x_i} = \sqrt{1} + \sqrt{2} + ... + \sqrt{20} \implies \sum_{i=1} ^{20}{x_i^2} = 1 + 2 + ... + 20\). Cauchy says sum of x^2 * sum of y^2 >= sum of (xy)^2. 

Please think this problem through, it is a basic problem that shouldn't require any help. Here are some hints: 

Note that each number is 6 plus the last number. So can we manipulate the sum to be divisible by six? That way we can apply the formula \(1 + 2 + ... + n = \frac{n(n-1)}{2}\)? 

Thank you for putting it in latex so its readable :) 

The basic idea when we see the word minimize is to think of inequalities. I suspect you got this problem from a book about inequalities. We also see squares, and that gives us the idea to use Cauchy Schwarz inequality. As a recap, it states \(\left| \sum_{i=1}^na_ib_i \right|^2 \le \left(\sum_{i=1}^{n}|a_i^2| \right) \left( \sum_{i=1}^n |b_i^2| \right) \). See here: https://artofproblemsolving.com/wiki/index.php/Cauchy-Schwarz_Inequality?srsltid=AfmBOore-GDD9oAtWZDDSJ-URH43dfQd2atn26PRaLxAr01bGo2-r183 (aops wiki page). 

This indeed matches with minimizing squares. Now, the question becomes, what do we let \(a_i\) and \(b_i\) be? [Looking back on it, we shouldn't use \(a_i\)to avoid confusion, but oh well... our "\(a_i\)" here is refering to the Cauchy Schwarz inequality, and not to the problem.] Well, let's look at our coefficients. we have 1, 2, 3, ... , 20. Since we are squaring things in our formula, we probably want something in the form \(\sqrt{1}, \sqrt{2}, \sqrt{3}, ... , \sqrt{20} \). So if we have\(a_i = \sqrt{i} \implies \sum_{i=1} ^{20}{a_i} = \sqrt{1} + \sqrt{2} + ... + \sqrt{20} \implies \sum_{i=1} ^{20}{a_i^2} = 1 + 2 + ... + 20\). What about \(b_i\)? We have \(a_i\) in the form of some square roots, but when we want to minimize \(\left| \sum_{i=1}^na_ib_i \right|^2\), we have \(A_i\) terms and whole numbers (here, \(A_i\) stands for the real numbers given in the problem). So, we want a \(b_i\) such that when it is multiplied by \(a_i\), we get nice coefficients. It becomes obvious (at least to me), that \( \sum_{i=1} ^{20}{b_i} = \sqrt{1} A_1 + \sqrt{2}A_2 + \sqrt{3}A_3 ... + \sqrt{20} A_{20} \implies \sum_{i=1} ^{20}{b_i ^ 2} = 1 A_1^2 + 2A_2^2 + 3A_3^2 ... + 20 A_{20}^2. \) Then, \( \sum_{i=1} ^{20}{(a_i b_i)^2} = 1A_1 + 2A_2 + 3A_3 ... + 20 A_{20}\). 

Finally, putting it all together, we have \(1A_1 + 2A_2 + 3A_3 + ... 20A_{20} \le (1 + 2 + 3 + ... + 20) (1A_1^2 + 2A_2^2 + 3A_3 ^ 2 + ... + 20 A_{20}^2)\). From the problem, we have \(1A_1 + 2A_2 + 3A_3 + ... 20A_{20} = 1\), and we know the sum of \((1 + 2 + 3 + ... + 20) = 210\)(You can think of it as grouping: 1+20 = 21, 2+19 = 21, 3+18 = 21. We have 10 total pairs, so 21x10 = 210. This is the derivation of the formula: \(1 + 2 + ... + n = \frac{n(n-1)}{2}\)). ]

Plugging these values in, we have \(\frac{1}{210} \le (1A_1^2 + 2A_2^2 + 3A_3 ^ 2 + ... + 20 A_{20}^2)\). 

Now, to minimize this, we need equality to hold. The equality condition is as follows: 

We need some \(x\) such that \(x a_i = b_i\), or \(x \sqrt{i} = \sqrt{i} A_i\), so \(x = A_i\) for all \(i \in {(1, 2, 3, ..., 20)}\)[The "e" notation is reads "in", so for all i "in" (1, 2, ..., 20).] 

From the original problem, we have \(1A_1 + 2A_2 + 3A_3 + ... 20A_{20} = 1\), so plugging in \(x = A_i\) gives \(1 + 2x + 3x + ... 20 x = 1 \implies x(1+2+3 + ... + 20) = 1, \text{so } x = \frac{1}{120}\). So, \(a_{12} = 12 \cdot \frac{1}{120} = \frac{1}{10}\). 

Note: I may have made some mistakes on the way, so make sure to read the solution, understand everything, and see what mistakes I made... Because just getting the answer without learning doesn't help you  

Thank you for trying, but you're answer is not correct. The correct answer was 0.913.

We can use complementary counting to solve this problem, by subtracting the probabilty that no adjacent faces are the same color:

1: All faces are colored diffferently. 6!=720 ways

2: One pair of opposite faces are colored the same. 3 ways to choose the pair of faces * 6 ways to color the faces * 120 ways to color the remaining faces = 2160 ways.

3: Two pairs of opposite faces are colored the same. 3 ways to choose the pairs of faces * 30 ways to color the faces * 12 ways to color the remaining faces = 1080 ways.

4: Each pair of opposite faces are the same. 6 * 5 * 4 ways to choose the colors = 120 ways.

(720 + 2160 + 1080 + 120) / 6^6 (total number of ways) ~ 0.08744.

1 - 0.08744 ~ 0.913.

Thank you for trying, and I hope this helped!

Let's solve this problem step-by-step.

1. Arithmetic Sequence and Sum Formulas

Let the arithmetic sequence be a1​,a2​,a3​,… with first term a1​ and common difference d.

The sum of the first n terms, Sn​, is given by:

Sn​=2n​[2a1​+(n−1)d]

2. Given Information

S20​=51​

S10​=0

3. Setting up Equations

Using the sum formula, we can write:

S20​=220​[2a1​+(20−1)d]=10(2a1​+19d)=51​

S10​=210​[2a1​+(10−1)d]=5(2a1​+9d)=0

4. Solving the Equations

From 5(2a1​+9d)=0, we have:

2a1​+9d=0

2a1​=−9d

a1​=−29​d

From 10(2a1​+19d)=51​, we have:

2a1​+19d=501​

Substitute 2a1​=−9d into 2a1​+19d=501​:

−9d+19d=501​

10d=501​

d=5001​

Now, substitute d=5001​ into 2a1​=−9d:

2a1​=−9(5001​)

2a1​=−5009​

a1​=−10009​

5. Find S_70

We want to find S70​:

S70​=270​[2a1​+(70−1)d]

S70​=35[2a1​+69d]

Substitute a1​=−10009​ and d=5001​:

S70​=35[2(−10009​)+69(5001​)]

S70​=35[−100018​+1000138​]

S70​=35[1000120​]

S70​=35[10012​]

S70​=35[253​]

S70​=25105​=521​

Therefore, S70​=521​.

Let's break down this problem step-by-step.

1. Laverne's Counting Pattern

Laverne counts by 5's, starting with 2. Her sequence is:

2, 7, 12, 17, 22, 27, 32, ...

2. Shirley's Sum

Shirley sums the numbers Laverne says. Let's track the sums:

2

2 + 7 = 9

9 + 12 = 21

3. Finding the Trigger

Shirley runs screaming when the sum exceeds 20.

The sum 9 is less than 20.

The sum 21 is greater than 20.

4. The Trigger Number

The number that caused the sum to exceed 20 was 12.

Therefore, the number Laverne says that sends Shirley screaming and running is 12.

Let's solve this equation step-by-step.

1. Substitution

To simplify the equation, let's make the following substitutions:

u = x - 4

a = x - 3 = u + 1

b = x - 5 = u - 1

The equation becomes:

a⁴ + b⁴ = -8 + 6ab³ - 11a³b

2. Rewrite the Equation

Substitute a = u + 1 and b = u - 1:

(u + 1)⁴ + (u - 1)⁴ = -8 + 6(u + 1)(u - 1)³ - 11(u + 1)³(u - 1)

Expand the terms:

(u⁴ + 4u³ + 6u² + 4u + 1) + (u⁴ - 4u³ + 6u² - 4u + 1) = -8 + 6(u + 1)(u³ - 3u² + 3u - 1) - 11(u³ + 3u² + 3u + 1)(u - 1)

2u⁴ + 12u² + 2 = -8 + 6(u⁴ - 2u³ + 0u² + 2u - 1) - 11(u⁴ + 2u³ - 2u - 1)

2u⁴ + 12u² + 2 = -8 + 6u⁴ - 12u³ + 12u - 6 - 11u⁴ - 22u³ + 22u + 11

2u⁴ + 12u² + 2 = -5u⁴ - 34u³ + 34u + -3

7u⁴ + 34u³ + 12u² - 34u + 5 = 0

3. Factorization

Let's try to find rational roots using the Rational Root Theorem. Possible rational roots are ±1, ±5, ±1/7, ±5/7.

By observation, u = 1/7 is not a root.

Let's check u = 1:

7 + 34 + 12 - 34 + 5 = 24 ≠ 0

Let's check u = -1:

7 - 34 + 12 + 34 + 5 = 24 ≠ 0

Let's check u = 5:

7(5⁴) + 34(5³) + 12(5²) - 34(5) + 5 ≠ 0

Let's check u = -5:

7(-5)⁴ + 34(-5)³ + 12(-5)² - 34(-5) + 5 ≠ 0

Let's check u = 1/7:

7(1/7)⁴ + 34(1/7)³ + 12(1/7)² - 34(1/7) + 5 ≠ 0

Let's check u = 5/7:

7(5/7)⁴ + 34(5/7)³ + 12(5/7)² - 34(5/7) + 5 ≠ 0

Let's check u = -5/7:

7(-5/7)⁴ + 34(-5/7)³ + 12(-5/7)² + 34(5/7) + 5 ≠ 0

Let's look at the equation again:

7u⁴ + 34u³ + 12u² - 34u + 5 = 0

Let's try to find a quadratic factor:

(7u² + au + 1)(u² + bu + 5) = 7u⁴ + (7b+a)u³ + (36+ab)u² + (5a+b)u + 5

Comparing coefficients:

7b + a = 34

36 + ab = 12

5a + b = -34

From 36 + ab = 12, ab = -24.

From 5a + b = -34, b = -34 - 5a.

Substitute into ab = -24:

a(-34 - 5a) = -24

-34a - 5a² = -24

5a² + 34a - 24 = 0

(5a - 4)(a + 6) = 0

a = 4/5 or a = -6

If a = 4/5, b = -34 - 5(4/5) = -34 - 4 = -38.

If a = -6, b = -34 - 5(-6) = -34 + 30 = -4.

Check 7b + a = 34:

7(-38) + 4/5 = -266 + 4/5 ≠ 34

7(-4) + (-6) = -28 - 6 = -34 ≠ 34

Let's try (7u² + au + 5)(u² + bu + 1) = 7u⁴ + (7b+a)u³ + (12+ab)u² + (a+5b)u + 5

7b+a = 34

12+ab=12

a+5b=-34

ab=0, so a or b is 0.

If a=0, 7b=34 so b is not integer.

If b=0, a=34 and a=-34, so impossible.

4. Numerical Solution

Using a numerical solver:

u ≈ -4.68977

u ≈ -2.44538

Thus:

x ≈ -0.68977

x ≈ 1.55462

5. Verification

Using the python code provided by Bard, the roots are approximately 2.44538 and 4.68977.

x ≈ 2.44538

x ≈ 4.68977

Final Answer: The final answer is 2.44538,4.68977​