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The fourth term of an arithmetic sequence is 141, and the seventh term is 132. The first term is _____.
 7 months ago
 7 months ago
The fourth term of an arithmetic sequence is 141, and the seventh term is 132. The first term is _____.
 7 months ago
 7 months ago

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satellite73Best ResponseYou've already chosen the best response.1
\[a, a+d,a+2d,a+3d,a+4d,...\] you got \(a+3d=141\) and \(a+6d=132\)
 7 months ago

satellite73Best ResponseYou've already chosen the best response.1
that means \(132141=(a+6d)(a+3d)=3d=9\) and so \(d=3\)
 7 months ago

satellite73Best ResponseYou've already chosen the best response.1
actually before we even start, since \(132<141\) is it clear that the terms are getting smaller?
 7 months ago

satellite73Best ResponseYou've already chosen the best response.1
in other words, "\(d\)" the "common difference" must be negative right?
 7 months ago

satellite73Best ResponseYou've already chosen the best response.1
if we call the first term \(a\) then the second term is \(a+d\) for some \(d\) and the third term is \(a+2d\), the fourth term is \(a+3d\) the fifth term is \(a+4d\) etc
 7 months ago

satellite73Best ResponseYou've already chosen the best response.1
in other words, you keep adding \(d\) to each term to get the next term, with the sophistication that you might be "adding" a negative number
 7 months ago

satellite73Best ResponseYou've already chosen the best response.1
The fourth term of an arithmetic sequence is 141 tells you that \(a+3d=141\)
 7 months ago

satellite73Best ResponseYou've already chosen the best response.1
you see that it is the fourth term, so it is \(a+3d\) not \(a+4d\)
 7 months ago

satellite73Best ResponseYou've already chosen the best response.1
and the seventh term is 132 means \[a+6d=132\]
 7 months ago

satellite73Best ResponseYou've already chosen the best response.1
from these two pieced of information we can solve for \(d\) and then solve for \(a\)
 7 months ago

satellite73Best ResponseYou've already chosen the best response.1
a bit of algebra shows that \[a+6d(a+3d)=3d\] right?
 7 months ago

satellite73Best ResponseYou've already chosen the best response.1
so we see that \[3d=132141=9\]
 7 months ago

lavalavaBest ResponseYou've already chosen the best response.1
and so since it is 74= 3 then 9/3 would be 3 right? giving us the difference
 7 months ago

satellite73Best ResponseYou've already chosen the best response.1
yeah \(3\) is the difference
 7 months ago

lavalavaBest ResponseYou've already chosen the best response.1
ohhh okay!!! i get it!! :D
 7 months ago

satellite73Best ResponseYou've already chosen the best response.1
you are still not done though right?
 7 months ago

satellite73Best ResponseYou've already chosen the best response.1
your question asked "The first term is _____"
 7 months ago

lavalavaBest ResponseYou've already chosen the best response.1
hmm well pluggin in the difference and then your equation... a4=141 a3=141+3=144 a2=144+3=147 a1=147+3=150 so then the first term would be 150 right? i just did it backwards
 7 months ago

satellite73Best ResponseYou've already chosen the best response.1
yeah i guess so i would have said \(a+3\times (3)=141\)or \[a9=141\] making \(a=150\) your method means you understand what is going on, which is good but you certainly wouldn't want to use that if you had say \(a_{75}\) and wanted \(a_1\)
 7 months ago

lavalavaBest ResponseYou've already chosen the best response.1
:D yay thank you!!! :D ohh okay... ill keep in mind that equation!! thank you soo much!
 7 months ago
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