1)The sum of 6 terms of an arithmetic series is 45, the sum of 12 terms is -8. Find the first term and the common difference?

2)If in a geometric sequence the sum of the second and the third is 20 and the sum of the fourth and fifth is 320, find the common ratio and the first term. Assume that the common ratio is possitive.

You have to know the formula for the sum of n terms of an AS

Sum(6) = 3(2a + 5d)
6a + 15d = 45 --- #1

sum(12) = 6(2a + 11d ) = -8
12a + 66d = -8
6a + 33d = -4 ---#2

subtract them
18d = -49
d = -49/18

back in #1
6a + 15(-49/18) = 45
6a = 45 + 735/18
6a = 515/18
a = 515/36

expected nicer numbers....

#2:

ar + ar^2 = 20 ---> ar(1 + r) = 20 ----#1
ar^3 + ar^4 = 320 ---> ar^3(1 + r) = 320 ---#2

divide #2 by #1
r^2 = 16
r = 4 , you said r is positive

in #1
4a(5) = 20
a = 1

a = 1 , r = 4

check:
terms would be
1 4 16 64 256
and 4+16 = 20
and 64 + 256 = 320

the common different between p and q

The third term of geometric sequence is 108 and the 6term is -32 find the common ratio ,first term and sum of first term

1) To find the first term and the common difference of an arithmetic series, we can use the formulas for the sum of the terms.

Let's start with the sum of the terms formula: Sn = n/2 * (2a + (n-1)d), where Sn is the sum of the first n terms, a is the first term, and d is the common difference.

Given that the sum of 6 terms is 45, we can substitute n = 6 and Sn = 45 into the formula:
45 = 6/2 * (2a + (6-1)d)
Simplifying this equation, we have:
45 = 3 * (2a + 5d)

Similarly, given that the sum of 12 terms is -8, we can substitute n = 12 and Sn = -8 into the formula:
-8 = 12/2 * (2a + (12-1)d)
Simplifying this equation, we have:
-8 = 6 * (2a + 11d)

Now we have a system of two equations:
45 = 3 * (2a + 5d)
-8 = 6 * (2a + 11d)

To solve this system, you can use any method you're comfortable with, such as substitution or elimination.

Let's use the elimination method. Multiply the second equation by 5 and the first equation by 6 to make the coefficients of d in both equations equal:
225 = 15 * (2a + 5d)
-40 = 30 * (2a + 11d)

Now subtract the second equation from the first equation to eliminate d:
265 = 2a - 25d

We now have a new equation:
265 = 2a - 25d

To find the values for a and d, we need one more equation. We can use either of the original equations. Let's use the first equation:

45 = 3 * (2a + 5d)

Simplify and rewrite this equation as:
15 = 2a + 5d

Now we have a system of two equations:
265 = 2a - 25d
15 = 2a + 5d

We can solve this system of equations using any appropriate method. Let's use the substitution method.

Rearrange the first equation to solve for a in terms of d:
2a = 265 + 25d
a = (265 + 25d) / 2

Substitute this value of a into the second equation:
15 = 2((265 + 25d) / 2) + 5d

Simplify and solve for d:
15 = 265 + 25d + 5d
15 - 265 = 30d
-250 = 30d
d = -250 / 30
d = -25/3

Now substitute this value of d back into the equation for a:
a = (265 + 25(-25/3)) / 2
a = (265 - 625/3) / 2
a = (795 - 625) / 6
a = 170 / 6
a = 85/3

Therefore, the first term (a) is 85/3 and the common difference (d) is -25/3.

2) To find the common ratio and the first term of a geometric sequence, we can use the formulas for the sum of the terms.

The sum of the terms in a geometric sequence is given by the formula: Sn = a * (r^n - 1) / (r - 1), where Sn is the sum of the first n terms, a is the first term, r is the common ratio, and n is the number of terms.

Given that the sum of the second and third terms is 20, we can substitute n = 3, Sn = 20 into the sum formula:
20 = a * (r^3 - 1) / (r - 1)

Similarly, given that the sum of the fourth and fifth terms is 320, we can substitute n = 5, Sn = 320 into the sum formula:
320 = a * (r^5 - 1) / (r - 1)

We now have a system of two equations:
20 = a * (r^3 - 1) / (r - 1)
320 = a * (r^5 - 1) / (r - 1)

To solve this system, we can eliminate a by multiplying the first equation by 16 and the second equation by 1:
320 = 16 * (r^3 - 1) / (r - 1)
320 = 16 * (r^5 - 1) / (r - 1)

Now we have a new system of equations:
320 = 16 * (r^3 - 1) / (r - 1)
320 = 16 * (r^5 - 1) / (r - 1)

Let's simplify both equations by multiplying both sides by (r - 1):
320 * (r - 1) = 16 * (r^3 - 1)
320 * (r - 1) = 16 * (r^5 - 1)

Simplifying further:
320r - 320 = 16r^3 - 16
320r - 16r^3 = 320 - 16

Rearrange the equation:
16r^3 - 320r + 304 = 0

Now we have a cubic equation in terms of r. To solve this equation, you can use various methods such as factoring, graphing, or using the cubic formula. After finding the value(s) for r, you can substitute it into either of the original equations to find the first term (a).