The first four terms of an arithmetic sequence are

2, x-y, 2x+y+7 and x-3y where x and y are constants.

Find the x and y and hence find the sum of the first 30 terms.

To find the values of x and y, we can use the fact that the terms in an arithmetic sequence have a common difference.

The common difference can be found by subtracting any two consecutive terms. Let's subtract the second term from the first term:

(x - y) - 2 = x - y - 2

Now, let's subtract the third term from the second term:

(2x + y + 7) - (x - y) = x + 2y + 7

Finally, let's subtract the fourth term from the third term:

(x - 3y) - (2x + y + 7) = -x - 4y - 7

Since these differences should all be equal, we can set up a system of equations:

x - y - 2 = x + 2y + 7
x + 2y + 7 = -x - 4y - 7

Simplifying these equations, we get:

-y - 2 = 2y + 7
x + 6y = -14

Solving the first equation for y:

3y = -9
y = -3

Substituting this value into the second equation:

x + 6(-3) = -14
x - 18 = -14
x = -14 + 18
x = 4

Therefore, x = 4 and y = -3.

To find the sum of the first 30 terms, we can use the formula for the sum of an arithmetic sequence:

Sn = (n/2)(2a + (n-1)d)

In this case, the first term (a) is 2, the common difference (d) is x - y, and n is 30.

Sn = (30/2)(2(2) + (30-1)(4-(-3)))

Simplifying,

Sn = 15(4 + 29(7))

Sn = 15(4 + 203)

Sn = 15(207)

Sn = 3105

Therefore, the sum of the first 30 terms is 3105.

To find the values of x and y, we can use the given terms in the arithmetic sequence.

The formula for the nth term of an arithmetic sequence is given by:

an = a1 + (n-1)d

Where:
an is the nth term of the sequence
a1 is the first term of the sequence
n is the position of the term in the sequence
d is the common difference between consecutive terms

Using this formula, we can start by finding the common difference between consecutive terms.

Given the first four terms:
a1 = 2
a2 = x - y
a3 = 2x + y + 7
a4 = x - 3y

The common difference between consecutive terms can be found by subtracting any two consecutive terms. Let's subtract a2 from a1:

a2 - a1 = (x - y) - 2 = x - y - 2

Since this represents the common difference, we can set it equal to the difference between a3 and a2 as well:

a3 - a2 = (2x + y + 7) - (x - y) = 2x + y + 7 - x + y = x + 2y + 7

Setting the two expressions equal to each other:

x - y - 2 = x + 2y + 7

Rearranging and simplifying:

y = -4

Substituting this value of y back into one of the equations, let's use a2:

x - (-4) - 2 = x + 8 - 2 = x + 6

Setting x + 6 equal to 2:

x + 6 = 2

Subtracting 6 from both sides:

x = -4

Hence, the values of x and y are x = -4 and y = -4.

To find the sum of the first 30 terms, we can use the formula for the sum of an arithmetic sequence:

Sn = (n/2)(2a1 + (n-1)d)

Where:
Sn is the sum of the first n terms
a1 is the first term of the sequence
n is the number of terms in the sequence
d is the common difference between consecutive terms

Substituting the values we have:

a1 = 2
n = 30
d = x - y - 2 = -6

Sn = (30/2)(2(2) + (30-1)(-6))
= 15(4 + 174)
= 15(178)
= 2670

Hence, the sum of the first 30 terms is 2670.

(2x+y+7) - (x-y) = (x-y)-2

(x-3y)-(2x+y+7) = (x-y)-2

x=2, y=-3

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