In an arithmetic sequence, the 7th term is 30, and the 11th term is 60. What is the 21st term of this sequence?

We can solve this system....where a1 is the first term and d is the common difference between terms

30   =  a1 + 6d         

60  =  a1  + 10d

Subtract the 1st equation from the 2nd

30  =  4d       divide both sides by 4

7.5  =  d

And using the second equation to find a1, we have

60  =  a1  + 10(7.5)

60  = a1  +  75           subtract 75 from both sides

-15    =  a1

So......the 21st term is

-15 + 7.5(20)  =

135

In an arithmetic sequence, the 7th term is 30, and the 11th term is 60. What is the 21st term of this sequence?

11 - 7 + 1 =5 number of terms

60 - 30 =30 - the difference between above 5 terms.

30 / 5 = 6 - the common difference. The nth. term is given by:

F + (N - 1) x D =nth. term, where F=first term, N=number of terms, D=common difference.

First term =30 - (6*6) = - 6

-6 + (21 -1) x 6 =

-6 + 20 x 6 =

-6 + 120 = 114 - the 21st. term.

P.S. There is something wrong with your sequence as stated. Either the 7th term or 11th term is wrong!. If the 7th term is right, then the 11th term is 54 and the 12th term is 60. If the 11th term is 60, then the 7th term should be 36. Either way, if the 1st term is 0, then the 7th should be 36 and the 11th should be 60.

If the 1st term is -6, then the 7th term should be 30 and the 11th term should be 54.

Check your sequence carefully.