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''
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.
the 20th term is 78
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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''
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''
Um... i get 66, but it's summer so i have no idea if thats right.


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

so T1+3*13=2(T1+3*5)

then for T20 its T1+3*19
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
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.