Arithmetic Operations

We need to find two numbers, let's call them x and y, that satisfy two conditions: their sum is 4 and their product is 2.

We can set up a system of two equations:
1) x + y = 4
2) xy = 2
We can solve this system by substitution, which will lead to a quadratic equation.

From the first equation, we can express y in terms of x:
y = 4 - x Now substitute this expression for y into the second equation:
x(4 - x) = 2 Expand the equation:
4x - x² = 2 Rearrange into the standard quadratic form ax² + bx + c = 0:
x² - 4x + 2 = 0 This equation cannot be easily factored, so we use the quadratic formula to solve for x:
x = -b ± √(b² - 4ac)2a Here, a=1, b=-4, and c=2.
x = -(-4) ± √((-4)² - 4(1)(2))2(1) x = 4 ± √(16 - 8)2 x = 4 ± √(8)2 Simplify the square root: √(8) = √(4 × 2) = 2√(2).
x = 4 ± 2√(2)2 Divide both terms in the numerator by 2:
x = 2 ± √(2) This gives us two possible values for x: x₁ = 2 + √(2) and x₂ = 2 - √(2).
If we take x = 2 + √(2), then y = 4 - x = 4 - (2 + √(2)) = 2 - √(2).
If we take x = 2 - √(2), then y = 4 - x = 4 - (2 - √(2)) = 2 + √(2).
In both cases, the two numbers are the same pair.

The two numbers are 2 + √(2) and 2 - √(2).

We need to list the natural numbers that, when divided by 4, give a remainder of 1.

A number that leaves a remainder of 1 when divided by 4 can be expressed in the form 4k + 1, where k is a non-negative integer (0, 1, 2, ...).

Let's find the first few terms of the sequence by substituting values for k:
- If k = 0, the number is 4(0) + 1 = 1. (1 divided by 4 is 0 with a remainder of 1).
- If k = 1, the number is 4(1) + 1 = 5. (5 divided by 4 is 1 with a remainder of 1).
- If k = 2, the number is 4(2) + 1 = 9. (9 divided by 4 is 2 with a remainder of 1).
- If k = 3, the number is 4(3) + 1 = 13. (13 divided by 4 is 3 with a remainder of 1).
The sequence continues in this pattern.

The sequence is 1, 5, 9, 13, ....

We need to determine if the sequence 1, 5, 9, 13, ... is an arithmetic sequence and provide a reason.

An arithmetic sequence is a sequence of numbers such that the difference between the consecutive terms is constant. This constant difference is called the common difference (d).

Let's check the difference between consecutive terms in the sequence:
- Difference between the 2nd and 1st term: 5 - 1 = 4.
- Difference between the 3rd and 2nd term: 9 - 5 = 4.
- Difference between the 4th and 3rd term: 13 - 9 = 4.
Since the difference is always 4, there is a constant common difference.

Yes, this is an arithmetic sequence. The reason is that there is a common difference of 4 between any two consecutive terms.

We need to create a new sequence by adding 2 to every term of the original sequence (1, 5, 9, 13, ...) and then find the algebraic formula for this new sequence.

The algebraic form (n-th term) of an arithmetic sequence is aₙ = a + (n-1)d. We can either find the formula for the original sequence and add 2, or we can find the new sequence first and then derive its formula.

Method 1: Find the new sequence first.
Original sequence: 1, 5, 9, 13, ...
Add 2 to each term:
New sequence: (1+2), (5+2), (9+2), (13+2), ...
New sequence: 3, 7, 11, 15, ...
For this new sequence:
- First term, a = 3.
- Common difference, d = 7 - 3 = 4.
Now, find its algebraic form:
aₙ = a + (n-1)d aₙ = 3 + (n-1)4 aₙ = 3 + 4n - 4 aₙ = 4n - 1 Method 2: Modify the algebraic form of the original sequence.
The algebraic form of the original sequence (1, 5, 9, ...) is xₙ = 1 + (n-1)4 = 1 + 4n - 4 = 4n - 3.
The new sequence aₙ is obtained by adding 2 to each term of xₙ.
aₙ = xₙ + 2 aₙ = (4n - 3) + 2 aₙ = 4n - 1 Both methods yield the same result.

The algebraic form of the new sequence is 4n - 1.

Using the same AP where a₁ + a₂₅ = 70, we need to find the 13th term (a₁₃).

From the previous part, we established that a₁ + a₂₅ = 2a + 24d = 70.
The 13th term is given by the formula a₁₃ = a + (13-1)d = a + 12d.
We can relate these two expressions.

We have the equation from the given information:
a₁ + a₂₅ = 70 2a + 24d = 70 We can factor out a 2 from the left side of the equation:
2(a + 12d) = 70 Now, divide both sides by 2:
a + 12d = (70)/(2) = 35 The expression for the 13th term is a₁₃ = a + 12d.
Therefore, a₁₃ = 35.

The 13th term of the sequence is 35.

Using the same AP, we need to find the sum of the first 25 terms (S₂₅).

There are two common formulas for the sum of an AP:
1. Sₙ = (n)/(2)[2a + (n-1)d]
2. Sₙ = (n)/(2)[a₁ + aₙ]
The second formula is more direct here since we are given information about a₁ and a₂₅.

We need to find S₂₅.
Using the formula Sₙ = (n)/(2)[a₁ + aₙ] with n=25:
S₂₅ = (25)/(2)[a₁ + a₂₅] We are given that a₁ + a₂₅ = 70.
Substituting this value into the sum formula:
S₂₅ = (25)/(2)(70) S₂₅ = 25 × 35 To calculate 25 × 35:
25 × 35 = 25 × (30 + 5) = (25 × 30) + (25 × 5) = 750 + 125 = 875 So, the sum of the first 25 terms is 875.

Alternative method using a₁₃:
We know the middle term is a₁₃ = 35. For an odd number of terms, the sum is the number of terms multiplied by the middle term.
S₂₅ = 25 × a₁₃ S₂₅ = 25 × 35 = 875 The sum of the first 25 terms is 875.

We are given two terms of an arithmetic sequence and asked to find the 15th term.
Given: a₇ = 100 and a₁₁ = 140.

The n-th term of an arithmetic sequence is aₙ = a + (n-1)d. We can set up a system of two linear equations with two variables, a (the first term) and d (the common difference), and solve for them.

Using the formula aₙ = a + (n-1)d, we can write the given information as two equations:
For the 7th term: a₇ = a + (7-1)d = a + 6d = 100 --- (1)
For the 11th term: a₁₁ = a + (11-1)d = a + 10d = 140 --- (2)
To find d, we can subtract equation (1) from equation (2):
(a + 10d) - (a + 6d) = 140 - 100 4d = 40 d = (40)/(4) = 10 The common difference is 10.
Now, substitute d=10 back into equation (1) to find a:
a + 6(10) = 100 a + 60 = 100 a = 100 - 60 = 40 The first term is 40.
Now we can find the 15th term (a₁₅):
a₁₅ = a + (15-1)d a₁₅ = 40 + 14(10) a₁₅ = 40 + 140 a₁₅ = 180 The fifteenth term of the sequence is 180.

Using the same AP as before (S₃=30, S₇=140), we need to write out the first three terms (a₁, a₂, a₃).

We have already found the constituent parts of the sequence:
- First term a = 5.
- Common difference d = 5.
- Second term a₂ = 10.
The terms of an AP are a, a+d, a+2d, ....

From the previous parts, we determined:
The first term a₁ = a = 5.
The second term a₂ = a + d = 5 + 5 = 10.
The third term a₃ can be calculated as a₂ + d.
a₃ = 10 + 5 = 15 Alternatively, using the formula:
a₃ = a + 2d = 5 + 2(5) = 5 + 10 = 15 So the first three terms are 5, 10, and 15.
We can check our work. The sum of these three terms is 5 + 10 + 15 = 30, which matches the given S₃ = 30.

The first three terms of the sequence are 5, 10, 15.

We are given the sum of the 1st and 25th terms of an AP. We need to find the sum of the 4th and 22nd terms.
Given: a₁ + a₂₅ = 70.

A key property of an arithmetic progression is that if the sum of the indices of two pairs of terms is equal, then the sum of the terms in those pairs is also equal.
That is, if p + q = r + s, then ap + aq = aᵣ + aₛ.

Method 1: Using the property.
We are given the sum of the 1st and 25th terms. The sum of their indices is 1 + 25 = 26.
We need to find the sum of the 4th and 22nd terms. The sum of their indices is 4 + 22 = 26.
Since the sum of the indices is the same (1+25 = 4+22), the sum of the terms must also be the same.
Therefore, a₄ + a₂₂ = a₁ + a₂₅ = 70.

Method 2: Using the general formula.
Let a be the first term and d be the common difference.
Given: a₁ + a₂₅ = 70.
a + (a + (25-1)d) = 70 a + a + 24d = 70 2a + 24d = 70 --- (1)
Now, let's find the sum of the 4th and 22nd terms:
a₄ + a₂₂ = (a + (4-1)d) + (a + (22-1)d)
a₄ + a₂₂ = (a + 3d) + (a + 21d)
a₄ + a₂₂ = 2a + 24d
From equation (1), we know that 2a + 24d = 70.
Therefore, the sum of the 4th and 22nd terms is 70.

The sum of the 4th and 22nd terms is 70.

We need to determine which of the given arithmetic sequences (APs) contains the number 37 as one of its terms.

The formula for the n-th term of an arithmetic sequence is aₙ = a + (n-1)d, where a is the first term, d is the common difference, and n is the term number.
For 37 to be a term in a sequence, the value of n calculated from the formula must be a positive integer.

Let's check each option:
(A) 4, 9, 14, ....
Here, a = 4 and d = 9 - 4 = 5.
Let aₙ = 37.
37 = 4 + (n-1)5 37 - 4 = (n-1)5 33 = 5(n-1) n-1 = (33)/(5) Since (33)/(5) is not an integer, n will not be an integer. So, 37 is not a term in this sequence.

(B) 7, 12, 17, ....
Here, a = 7 and d = 12 - 7 = 5.
Let aₙ = 37.
37 = 7 + (n-1)5 37 - 7 = (n-1)5 30 = 5(n-1) n-1 = (30)/(5) = 6 n = 6 + 1 = 7 Since n = 7 is a positive integer, 37 is the 7th term of this sequence.

(C) 8, 12, 16, ....
Here, a = 8 and d = 12 - 8 = 4.
Let aₙ = 37.
37 = 8 + (n-1)4 29 = 4(n-1) n-1 = (29)/(4) Since (29)/(4) is not an integer, n will not be an integer. So, 37 is not a term in this sequence.

(D) 7, 11, 15, ....
Here, a = 7 and d = 11 - 7 = 4.
Let aₙ = 37.
37 = 7 + (n-1)4 30 = 4(n-1) n-1 = (30)/(4) = 7.5 Since 7.5 is not an integer, n will not be an integer. So, 37 is not a term in this sequence.

The only sequence for which n is a positive integer is 7, 12, 17, .... Therefore, 37 is a term of this sequence.

We need to find the algebraic expression for the n-th term (aₙ) of the given arithmetic sequence.

The formula for the n-th term of an arithmetic sequence (AP) is:
aₙ = a + (n-1)d where a is the first term and d is the common difference.

The given arithmetic sequence is 6, 10, 14, ....
The first term is a = 6.
The common difference is d = 10 - 6 = 4. (Also, 14 - 10 = 4).
Now, we substitute these values into the formula for the n-th term:
aₙ = 6 + (n-1)4 Distribute the 4:
aₙ = 6 + 4n - 4 Combine the constant terms:
aₙ = 4n + 2 The algebraic form of the sequence is 4n + 2.