## yummydum Group Title Using sigma notation, describe the total length of 7 pictures hung side-by-side. Then find the total length the pictures occupy along the wall if the first picture has a length of 10 inches and the length of each successive picture is 4 inches longer than the previous one. one year ago one year ago

1. yummydum

@hartnn

2. hartnn

let l1 to l7 be length of pics. from n=1 to 7, l_n and sum notation : $$\sum \limits_{n=1}^7l_n$$ is your 1st part.

3. ParthKohli

Hint: for the first part, denote the length of the seven pictures by a sequence $$a_1, a_2 \cdots a_7$$

4. hartnn

l1 =10 and common difference d =4

5. yummydum

rightI got that but for the equation..?

6. hartnn

you need to find n'th tern l_n with l_1 =10 , d = 4 $$l_n = l_1+(n-1)d$$

7. yummydum

10, 14, 18,..$a_n=(10)-(n-1)(4)$$a_n=10-4n+4$$a_n=14-4n$$\sum_{n=1}^{7}~14n-4$yes?

8. hartnn

middle sign is + ln = 6+4n

9. yummydum

oh ...i knew that..

10. hartnn

now can you solve the summation ?

11. hartnn

$$\sum 6+4\sum n$$

12. ParthKohli

The worst pain, after the famines and diseases, is typing on an iPad. :-(

13. hartnn

know whats $$\sum n=..?$$

14. yummydum

wow it wrote that weird.. $\large\sum_{n=1}^{7} ~6+4n~~ \implies~~154$

15. ParthKohli

I don't know, just pointing out that a common mistake students sometimes make is that they do $$\sum 6 = 6$$ when it's not.

16. hartnn

i also get 154

17. ParthKohli

You didn't make that mistake though, nice :-)

18. yummydum

err..what? i did $$[6+(4\times1)]+[6+(4\times2)]+[6+(4\times3)]+$$$$[6+(4\times4)]+[6+(4\times5)]+[6+(4\times6)]+[6+(4\times7)]$$

19. ParthKohli

You are right.

20. yummydum

10+14+18+22+26+30+34=154.....yup :)

21. ParthKohli

I never said you were wrong lol, just pointing out a common mistake.

22. ParthKohli

Good stuff :-)

23. yummydum

lol, thanks guys!

24. yummydum

whats 8(n+3)??

25. yummydum

10+14+18+22+26+30+34=154 ....is that not right?

26. yummydum

o.O

27. yummydum

lol no probs