## anonymous 5 years ago please help The 11th term of an arithmetic progression is 1. the sum of the first 10 terms is 120. find the 4th term :S formula for the... sum of terms=1/2n(a+l)=1/2n{2a+(n-1)d} thanks

1. Yuki

okay

2. Yuki

${11(a_1 + a_{11}) \over 2} = 120 + 1$

3. Yuki

so${a_1 + 1} = 22$ thus a_1 = 21

4. Yuki

Then you can say$a_{11} = a_1 + 10d$ $a_11 = 1 = 21 + 10d$

5. Yuki

so d = -2

6. Yuki

that means that the 4th term is 21+3(-2) = 15

7. Yuki

:)

8. Yuki

did it make sense ?

9. anonymous

but you didnt even use 4 once :S??? and i dont really understand where 21+3(-2)=15 comes from :(

10. Yuki

ok, $a_n = a_1 + d(n-1)$ so$a_4 = a_1 + (-2)(4-1) = 21 + 3(-2)$

11. Yuki

the idea of arithmetic sequence is that you are "Adding the same number for all sequence"

12. anonymous

thanks :)

13. Yuki

np :)

14. anonymous

sorry but i first thought that i knew where 11(a1+a11)2=120+1 came from but i dont :S

15. Yuki