In an arithmetic progression that consists of 60 limits overall
- The sum of first 20 limits is 200
- The sum of the last limits is 1000
What's the sum of all the 60 limits?

The rule you should use is
Sn=n/2(2a1+(n-1)d)

In an arithmetic progression that consists of 60 limits overall
by "limits", do you mean the number of terms of the arithmetic series?

- The sum of first 20 limits is 200
- The sum of the last limits is 1000
I understand the first statement to be $a_1 + a_2 + ... + a_{20} = 200$

does the second statement mean $a_{21} + a_{22} + a_{23} + ... + a_{60} = 1000$ ?

What's the sum of all the 60 limits?
shouldn't $S_{60} = 1200$

... please provide a correction if there is a misunderstanding.

Originally Posted by skeeter
by "limits", do you mean the number of terms of the arithmetic series?

I understand the first statement to be $a_1 + a_2 + ... + a_{20} = 200$

does the second statement mean $a_{21} + a_{22} + a_{23} + ... + a_{60} = 1000$ ?

shouldn't $S_{60} = 1200$

... please provide a correction if there is a misunderstanding.
You're right, sorry English isn't my first language.
But no unfortunately the answer is 1800

$S_{20} = \dfrac{20}{2}[2a_1 + (20-1)d] = 200$

- The sum of the last limits is 1000
maybe the last 20 terms? ...

$S_{60}-S_{40} = \dfrac{60}{2}[2a_1 + (60-1)d] - \dfrac{40}{2}[2a_1 + (40-1)d]= 1000$

Originally Posted by skeeter
$S_{20} = \dfrac{20}{2}[2a_1 + (20-1)d] = 200$

maybe the last 20 terms? ...

$S_{60}-S_{40} = \dfrac{60}{2}[2a_1 + (60-1)d] - \dfrac{40}{2}[2a_1 + (40-1)d]= 1000$

Yes the last 20 sorry i forgot to type thaaat!

well, you have two equations with two unknowns, $a_1$ and $d$ ... solve the system.