#1
So as usuall im having trouble with math, anyways i need a help with an exercise, its about arithmetic series and sequences....... the problem says to find the last term in a sequence of 20 terms, in other words find the 20th term, and the numbers they give me are........... t(term) sub 14 = 2 times tsub6 and the common difference (d) is 3.......so yeah i need help and il reward you in any way possible pleasee help me out, my whole school year depends on this exam! please help

EDIT: Sorry i forgot to mention i would this with no problem if they stated the terms right, all i need help with is how to know the value of t14 or even t6.
''courage is not enough to forget''
#3
I would help, but I have absolutely no idea how I would attempt to solve that. Sorry, lol we don't learn that stuff in 8th grade math.
#4
the 20th term is 78
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#5
Quote by jimmy_neutron
what do you man by t sub 14??

have you ever seen on chemistry where for example they show a big C with a little number to the side?? that little number is the sub so itd be a big T with a little 14 to the side
''courage is not enough to forget''
#7
oh yeahh!!!! your right thanks awesome man!!!


-_-


im serious come on it cant be that hard, im just stupid for math
''courage is not enough to forget''
#8
Um... i get 66, but it's summer so i have no idea if thats right.

Basically,

set T14=2*T6
since the common difference is 3, that means each T sub n is T1 plus 3*(n-1).
so...
T14=T1+3*13
and T6=T1+3*5

so T1+3*13=2(T1+3*5)
T1+39=2T1+30
9=T1

then for T20 its T1+3*19
=9+3*19=66
#9
What i got for the 20th term was 66.

That should be right cuz then for a 6th term you get 24, and for the 14th term you get 48 which is twice as much. From there you can get the equation: A sub n=3n+3

I really think that's right...I hope so, i had to do a lot harder math than that this year
#10
Quote by Spaethor!
From there you can get the equation: A sub n=3n+3



That works too, but the equation is A sub n=3n+6.