# The sum of 12 terms of an arithmetic series is 186 and the 20th term is 83. Find the sum of 40 terms.

*print*Print*list*Cite

The previous answer has elements that are absolutely correct but the solution is unfortunately incorrect due to "mjripalda" misreading the fact that the second equation relates to T20 not S20.

The formula for the sum of an arithmetic sequence can be written:

`S n = n/2 ( 2a + (n-1)d)`

The formula for a term in an arithmetic sequence is :

`T n = a + (n - 1)d`

Therefore:

`S (12) = 12/2 (2a + 11d)`

`T20 = a + 19d`

We know that S 12 = 186 and we know that T20 = 83 and we do not know a (the first term) or d (the difference). So we use simultaneous equations to solve:

186= 6(2a + 11d)

`186/6 = 2a + 11d`

`31= 2a + 11d`

now concentrate on the other equation:

`83 = a + 19d`

to eliminate "a" multiply your second equation by 2:

166 = 2a + 38d

Now take it away from your first equation:

`(31 = 2a + 11d)`

`-(166 = 2a + 38d)`

Do vertical subtraction such that 31- 166 and so on:

- 135 = -27d

`therefore d= 5`

Revert to one of your previous equations to find a:

`83= a + 19 (5)`

`83 - 95 = a`

`therefore a = -12`

Now you are ready to find the sum to 40 terms:

`Sn = n/2 (2a + (n-1)d)`

`S40 = 40/2 ( 2 (-12) + (40-1) (5))`

`S40 = 20( -24) + 195)`

`S40 = 20 (171)`

**S40 = 3420**