determind the arithmetic sequence if fourth term is6 and the eleventh term is _34?

To determine the arithmetic sequence, we need to find the common difference (d) between the terms.

Let's use the formula for the nth term of an arithmetic sequence:

an = a1 + (n - 1)d

where:
an = nth term
a1 = first term
n = position of the term
d = common difference

We are given that the fourth term (a4) is 6 and that the eleventh term (a11) is -34. We can plug in these values into the formula.

For the fourth term:
a4 = a1 + (4 - 1)d
6 = a1 + 3d

For the eleventh term:
a11 = a1 + (11 - 1)d
-34 = a1 + 10d

Now we have a system of two equations:

6 = a1 + 3d ---(1)
-34 = a1 + 10d ---(2)

We can solve this system of equations to find the values of a1 and d.

Subtracting equation (1) from equation (2) gives:

-34 - 6 = (a1 + 10d) - (a1 + 3d)
-40 = 7d

Dividing both sides of the equation by 7 gives:

7d/7 = -40/7
d = -40/7

Now we can substitute the value of d back into equation (1) or (2) to find a1.

Let's use equation (1):

6 = a1 + 3(-40/7)
6 = a1 - 120/7

To simplify, let's convert 6 to a fraction with a common denominator:

6 = 42/7

Now, we can rewrite the equation:

42/7 = a1 - 120/7

Adding 120/7 to both sides:

42/7 +120/7 = a1
162/7 = a1

Therefore, the first term (a1) of the arithmetic sequence is 162/7, and the common difference (d) is -40/7.

For your AP

term4 = a+3d = 6 **
term11=a+10d = 34 ***
subtract them
7d = 28
d = 4
sub in **
a+12 = 6
a = -6

the AP is -6, -2, 2, 6, 10, ...