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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.

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

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Other answers:

l1 =10 and common difference d =4
rightI got that but for the equation..?
you need to find n'th tern l_n with l_1 =10 , d = 4 \(l_n = l_1+(n-1)d\)
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?
middle sign is + ln = 6+4n
oh ...i knew that..
now can you solve the summation ?
\(\sum 6+4\sum n\)
The worst pain, after the famines and diseases, is typing on an iPad. :-(
know whats \(\sum n=..?\)
wow it wrote that weird.. \[\large\sum_{n=1}^{7} ~6+4n~~ \implies~~154\]
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.
i also get 154
You didn't make that mistake though, nice :-)
err..what? i did \([6+(4\times1)]+[6+(4\times2)]+[6+(4\times3)]+\)\([6+(4\times4)]+[6+(4\times5)]+[6+(4\times6)]+[6+(4\times7)]\)
You are right.
10+14+18+22+26+30+34=154.....yup :)
I never said you were wrong lol, just pointing out a common mistake.
Good stuff :-)
lol, thanks guys!
whats 8(n+3)??
10+14+18+22+26+30+34=154 ....is that not right?
o.O
lol no probs

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