Two terms of an arithmetic sequence are t6=-3 and t11=-13. What is t20?

t(6) = -3

t(11) = -13

-3 - (-13) = 10
10/5 = 2

Go one term back from last term to find term 1.

t(10) = 10(2) + -13 = 7

t(1) = 7
Now you can find term 0

7 + 2 = 9
t(0) = 9

Common difference is: d = -2

Equation: tn = -2n + 9

t(1) = 7
t(2) = 5
t(3) = 3
t(4) = 1
t(5) = -1
t(6) = -3
t(11) = -13
t(20) = -31

or, just note that T20 = T11+9d = -13-18=-31

Thank you

To find the value of t20 in the arithmetic sequence, we need to calculate the common difference (d) and then use it to find the 20th term.

The formula for the nth term (t_n) of an arithmetic sequence can be written as:

t_n = a + (n - 1)d

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

From the given information, we can determine that t6 = -3 and t11 = -13.

Using the formula, we can substitute the values into the equations:

t6 = a + (6 - 1)d = -3
t11 = a + (11 - 1)d = -13

Simplifying these equations, we get a + 5d = -3 and a + 10d = -13.

To solve for a and d, we can subtract the first equation from the second equation:

(a + 10d) - (a + 5d) = -13 - (-3)
10d - 5d = -10
5d = -10
d = -10 / 5
d = -2

Now that we know the value of d, we can substitute it back into one of the original equations to solve for a:

a + 5d = -3
a + 5(-2) = -3
a - 10 = -3
a = -3 + 10
a = 7

Now that we have found the values of a and d, we can find the 20th term (t20):

t20 = a + (20 - 1)d
t20 = 7 + 19(-2)
t20 = 7 - 38
t20 = -31

Therefore, the value of t20 in the arithmetic sequence is -31.