Question

given the arithmetic series 34+31+28+25+22+.....
Find the sum of the first n terms when n=32 ?

Answer 1

it appears that they are sequenced by -3 from the one before

Answer 2

what do you know about the formula for an arith.seq.?

Answer 3

well yea i figured d=-3 term1 was 34 and n=32 and the is sn=(t1+tn) n/2 but you need the last term to get it so you try tn=t1+ d(n-1) but i dont know how to work them out right d= common differance= -3 n= number of terms= 32

Answer 4

looks about right; let me try then

an = 34 -3(n-1)

a32 = 34 -3(31)

a32 = 34-93

a32 = -59 is what i get

Answer 5

thats just for the a32 to

Answer 6

to sum up; we can add the first and the last terms and multiply by how many terms/2

Answer 7

there is some ploy about the numbers being seperated by more than 1 tho

Answer 8

4+8+12+16 = 40

4+16 = 20*4/2 = 40.... then again, maybe not

Answer 9

um? even you can't solve the math problem i posted 0.0?

Answer 10

lol .... i tried to go to it and my browser froze, its on my queue tho

Answer 11

oh...o.k..sorry to distrub ya!

Answer 12

thank you so freiken much now i have to figure out how to do the sn=(t1+tn) n/2

Answer 13

34 + 31 + ... + -56 + -59
-59 -56 + ... + 31 + 34
-------------------------
-25 -25 - ... -25 -25

looks to be -25(32)/2

Answer 14

when in doubt, flip it out and do it the old school

Answer 15

it tells us that we are adding up -25, well, 32 times; but that is 2 times to many so half it

Answer 16

thanks man this is my first time ever on here and you made it worth it i appreciate it

Answer 17

youre welcome, and good luck :)

Answer 18

given the arithmetic series 34+31+28+25+22+.....
find N for Sn=-12
does anybody have a clue???? do you understand this?? i wouldnt even know what equation to use? please help me

Answer 19

given the arithmetic series 34+31+28+25+22+.....
find N for Sn=-12
does anybody have a clue???? do you understand this?? i wouldnt even know what equation to use? please help me

Answer 20

34 + x = -12

x = -46

-46 = 34 -3(n-1)

-46-34 = -3(n-1)

-80/-3 = n-1

80/3 + 1 = n

might not be what im thinking of tho

Answer 21

i know the answer i need to know how i got it like the right equation? the answer is 24

Answer 22

24 is the number of terms it takes to make the sum of the sequence -12 but i need to know how to do that does anybody have a clue?

Answer 23

yeah, your spose to solve for n in the equation

Answer 24

do you know the equation i need to know how to solve for n

Answer 25

-12 = 34 -3(n-1)

-12-34 = -3(n-1)

-46/-3 = n-1

15 .... it aint working out to a good number

Answer 26

lets see if its 24 for starts

a24 = 34 -3(23)

a24 = 34 - 69

a24 = -25

-12 is NOT the 24the term

Answer 27

are you using a different sequence now?

Answer 28

now the SUM of the nth terms is -12

Answer 29

yea i know i dont get it my teacher is crazy
no i dont have any more i just need to find n if the sum = -12 i have no clue what to do but by the way thank you for being patient with me and trying

Answer 30

-12 = (f+l)n/2

-24 = (34+l )n

ill never learn if i dont try :) we can brunt it out if its small enough

Answer 31

34 + (34-3(n-1)) n
----------------- = -12 is what we are gong for
2

Answer 32

i did already thats how i came up with 24? lol
i thought that equation was for geometric series????

Answer 33

34n + (34-3n-3)n = -24

34n + 31n -3n^2 = -24

0 = 3n^2 -65n -24, this might help if i stuck the the math right

Answer 34

nah, this equation states its arith seq.

Answer 35

"given the arithmetic series 34+31+28+25+.."

Answer 36

umm????? i am lost

Answer 37

i already see a mistake :)

71n -3n^2 = -24

Answer 38

lets start at the begining:

we know that if this is going to work, then the formula for the sum of a sequance will have to fit.
\[\frac{n}{2}(a1 + (a1+d(n-1)))=-12\]

Answer 39

we dont know what the "last" term is, but it will have to fit into that set up

Answer 40

yes??

Answer 41

\[\frac{n}{2}(a1 + (a1+d(n-1)))=-12\]
\[\frac{n(2)}{2}(a1 + (a1+d(n-1)))=-12(2)\]
\[n(a1 + (a1+dn-d))=-24\]
\[a1n + n(a1+dn-d)=-24\]
\[a1n + a1n+dn^2-dn=-24\]
\[2a1n+dn^2-dn=-24\]
you with it so far?

Answer 42

gettin there

Answer 43

\[2a1n+dn^2-dn=-24\]fill in the terms we know now
\[2(34)n+(-3)n^2-(-3)n=-24\]
\[68n-3n^2+3n=-24\]
\[71n-3n^2=-24\]
\[0=3n^2-71n-24\]

Answer 44

now our options for "n" will be whatever we get from the quadratic formula; disregarding any negative results

Answer 45

im not sure how much you are aware of the quad formula but here it is:
\[\frac{-b+\sqrt{b^2-4ac}}{2a};\text{ given }ax^2+bx+c=0\]

Answer 46

wow! never seen that one before lol

Answer 47

my teacher never went over this so on my test tommorow i have to do this??

Answer 48

\[0=3n^2-71n-24\]
\[n=\frac{71\pm\sqrt{71^2-4(3)(-24)}}{2(3)}\]
\[n=\frac{71\pm\sqrt{5041-12(-24)}}{6}\]
\[n=\frac{71\pm\sqrt{5041+288}}{6}\]
\[n=\frac{71\pm\sqrt{5329}}{6}\]
\[n=\frac{71\pm73}{6}\]
\[n=\frac{-2}{6},\frac{144}{6}\]
\[n=\cancel{\frac{-2}{6}},24\]

Answer 49

you have to if your gonna find the n yes

Answer 50

the other option is to try and err till you find something close and narrow it in

Answer 51

ummmm ...... i see something that i didnt before

Answer 52

so n=4 for the sum to be -12?

Answer 53

24 ... but look

remember up there how i got to this step?
\[n(a1 + (a1+dn-d))=-24\]
notice how |-24| = 24 ?

Answer 54

and how we want an answer of 24?

Answer 55

yea! ok im starting to click a little

Answer 56

34+31+28+25

\[\frac{4}{2}(34+25)=118\]
i think it was just a coincidence since 118(2) is NOT equal to 4

Answer 57

you just lost me again so is there a shorter way to show my work on this problem from the beggening?

Answer 58

\[\frac{n}{2}(34+[34-3(n-1)])=-12\]
we would still have to find n

Answer 59

we could do trial and error; start at 10 and work it out if need be.

\[\frac{n}{2}(34+[34-3(n-1)])=-12\]
\[n(34+34-3n+3)=-24\]
\[n(71-3n)=-24\]
\[10(71-3(10))=-24\]
\[10(71-30)=-24\]
\[10(41)=-24;not.it\]------------------------
\[20(71-3(20))=-24\]
\[20(11)=-24;not.it\]------------------------
\[30(71-3(30))=-24\]
\[30(-19)=-24\]
\[-570=-24\]
well, its between 20 and 30

Answer 60

damn so what is n 24?
was that right?

Answer 61

\[n(71-3n)=-24\]
\[25(71-3(25))=-24\]
\[25(-4)=-24\]
\[-100=-24\]----------------
\[24(71-3(24))=-24\]
\[24(71-72)=-24\]
\[24(-1)=-24\]
and we find it thru trial and error

Answer 62

ok

Answer 63

does the trial and error make sense?

Answer 64

no treally to me im a junior in high school no one ever tought me this she just said get n lol

Answer 65

well, i dont know how to make it any simpler than the trial and error.

Answer 66

we take what we know; the sum formula for arithmetic seq; and fill test out "n"s till we find one that sticks in a systematic manner

Answer 67

eventually it narrow down to the right choice

Answer 68

anywhos; n=24

a24 = 34 -3(23)

a24 = 34 -69

a24 = -35

Answer 69

thank you man i would be lost with out ya

Answer 70

youre welcome; it takes practice, and not really something you can just acquire over night :)

Answer 71

well the test is tommorow so i can acquire as much as i can lol