find the twenty-first term of an A.S of which the 6th term is 3 and the 14th is 19

a a+d a+2d a+3d .... a+(n-1)d

sixth term = a+5d = 3
fourteenth term = a+13d = 19
subtract to eliminate a
-8d = -16
d = 2
then a + 10 = 3
a = -7

twenty first term = -7 + 20(2)
= -7 + 40 = 33

19+(21-19)(19-3)/(14-6) = 23

Should have said

19+(21-14)(19-3)/(14-6) = 33

To find the twenty-first term of an arithmetic sequence, we need to determine the common difference and the first term.

First, let's find the common difference (d) using the given information. We're given that the 6th term (a6) is 3, and the 14th term (a14) is 19.

The formula to find the nth term of an arithmetic sequence is:

an = a1 + (n - 1) * d

Where:
an = nth term
a1 = first term
d = common difference

To find d, we can create two equations using the given information:

a6 = a1 + (6 - 1) * d (Equation 1: 6th term)
a14 = a1 + (14 - 1) * d (Equation 2: 14th term)

Using the first equation, we can substitute the values:

3 = a1 + 5d

Using the second equation, we can substitute the values:

19 = a1 + 13d

We now have a system of two equations:

3 = a1 + 5d (Equation 3)
19 = a1 + 13d (Equation 4)

Now, subtract Equation 3 from Equation 4 to eliminate a1:

19 - 3 = (a1 + 13d) - (a1 + 5d)
16 = 8d

Dividing both sides by 8:

d = 2

Now, we have found the common difference (d = 2).

To find the first term (a1), we can choose either Equation 3 or Equation 4 and substitute the value of d:

3 = a1 + 5(2)
3 = a1 + 10
a1 = -7

Now we have found the first term (a1 = -7) and the common difference (d = 2).

Finally, we can find the twenty-first term (a21) using the formula for the nth term:

a21 = a1 + (21 - 1) * d
a21 = -7 + 20 * 2
a21 = -7 + 40
a21 = 33

Therefore, the twenty-first term of the arithmetic sequence is 33.