what is the sum of the arithmetic sequence 135,123,111... if there are 34 terms?
Still Need Help?
Join the QuestionCove community and study together with friends!
Sign Up
OpenStudy (amistre64):
what is the difference between the 2nd term and the 1st term?
OpenStudy (amistre64):
123 - 135 = ?
OpenStudy (anonymous):
12.
OpenStudy (anonymous):
Dude ,Try to solve your own exam by your own ,believe me.....u gonna solve it faster.
OpenStudy (amistre64):
lets go with -12 :) this is our common difference (d).
Still Need Help?
Join the QuestionCove community and study together with friends!
Sign Up
OpenStudy (anonymous):
ok.
OpenStudy (amistre64):
we can use the formula:
an = a1 + d(n-1)
to find the last term that we will need
OpenStudy (amistre64):
when n=34 we get:
a34 = 135 -12(33)
whats tat get us?
OpenStudy (anonymous):
a34= 135- 396?
OpenStudy (amistre64):
keep going, so far so good
Still Need Help?
Join the QuestionCove community and study together with friends!
Sign Up
OpenStudy (anonymous):
i dont kow how to keep going.
OpenStudy (amistre64):
135 - 396 = ??
OpenStudy (anonymous):
-261
OpenStudy (amistre64):
good; this is our last term in the sequece that we need
OpenStudy (amistre64):
now we need to use another nifty little formula:
\[Sn=\frac{n}{2}(a1+an)\]
Still Need Help?
Join the QuestionCove community and study together with friends!
Sign Up
OpenStudy (amistre64):
\[S_{34}=\frac{34}{2}(135-261)\]
OpenStudy (anonymous):
@amistre64 where can I find proof for that formula?
OpenStudy (anonymous):
okay... now?
OpenStudy (amistre64):
now thats the answer
OpenStudy (anonymous):
no its not. its not an option.
Still Need Help?
Join the QuestionCove community and study together with friends!
Sign Up
OpenStudy (amistre64):
the proof is pretty simple xcrypt
suppose we have an arith seq: a1 to an
OpenStudy (amistre64):
try doing the math to finish it out; im not your nanny
OpenStudy (anonymous):
i know its a negative.
OpenStudy (anonymous):
so is it like, -2,108?
OpenStudy (amistre64):
\[A = a_1+a_2+a_3+...+a_n\]
to sum this up we can add it together twice, but in a useful way, reverse the second set
\[A = a_1+a_2+a_3+...+a_n\]\[A = a_n+a_{n-1}+a_{n-2}+...+a_1\]
that gets us
2a = (a1+an) + (a1+an) + ... + (a1+an) ; n times
Still Need Help?
Join the QuestionCove community and study together with friends!
Sign Up
OpenStudy (anonymous):
was my answer correct?
OpenStudy (anonymous):
@Ethan96 use a calculator? Or just type in the equation in google
OpenStudy (amistre64):
17*(-126) i believe
OpenStudy (anonymous):
-2,142
OpenStudy (amistre64):
divide both sides by 2 to get:
\[A=\frac{n}{2}(a_1+a_n)\]
Still Need Help?
Join the QuestionCove community and study together with friends!
Sign Up
OpenStudy (anonymous):
okay,so -2,142 is correct? :D
OpenStudy (amistre64):
that looks better yes
OpenStudy (anonymous):
THANKS
OpenStudy (amistre64):
theres other proof, but that ones the one i like
OpenStudy (anonymous):
hmm great, thanks! What's this topic called in math? Never seen it before
Still Need Help?
Join the QuestionCove community and study together with friends!
Sign Up
OpenStudy (amistre64):
series and sequences is usually when it comes about, but that little tidbit also comes up in discrete math