Question

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.

Answer 1

@hartnn

Answer 2

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.

Answer 3

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

Answer 4

l1 =10 and common difference d =4

Answer 5

rightI got that but for the equation..?

Answer 6

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

Answer 7

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?

Answer 8

middle sign is +
ln = 6+4n

Answer 9

oh ...i knew that..

Answer 10

now can you solve the summation ?

Answer 11

\(\sum 6+4\sum n\)

Answer 12

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

Answer 13

know whats \(\sum n=..?\)

Answer 14

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

Answer 15

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.

Answer 16

i also get 154

Answer 17

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

Answer 18

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

Answer 19

You are right.

Answer 20

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

Answer 21

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

Answer 22

Good stuff :-)

Answer 23

lol, thanks guys!

Answer 24

whats 8(n+3)??

Answer 25

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

Answer 26

o.O

Answer 27

lol no probs