Question

I need help with:

The sum of the first 10 term of an Arithmetic.Progression is 120 and the sum of the first twenty terms is 840. Find the sum of the first thirty terms.

Answer 1

10(2a+9r)/2=120
20(2a+19r)/2=840
solve that and use a and r

Answer 2

quetejedi, shouldn't it be 5 and 10, the formula is N/2?

Answer 3

there is it, i divided by 2

Answer 4

\[\S_{n}=n/2(2a+(n-1)d)\]120=5(2a+9d)
840=10(2a+19d)
120=10a+45d
840=20a+15960d
240=20a+90d

840=20a+15960d

d=26.45
a=-107.025
gues this can help

Answer 5

Thank you very much, this helped me.

I think you have 840=20a+15960d wrong it should be 190 because of
840=10(2a+19d)

Answer 6

I did it like this:
S10 = 5 ( 2a + 9d) = 120
S20 = 10 (2a + 19d) = 840

2a +9d = 120/5 = 24
2a + 19d = 840 / 10 = 84

then 10d (from 19-9d) = 84 + 24 = 60

d=6 a = -15

530 = 15 (2* - 15 + 29 * 6 = 2160?

Answer 7

yap