Question

For an arithmetic series, given s12=186 and t20=83, what is s40?

Answer 1

\[\frac{12(t_1+t_{12})}{2}=186\]
\[t_{20} = 83\]
\[\frac{40(t_1+t_{40})}{2}=\_\_\]

Answer 2

or do we start at a t0?

Answer 3

186/6 = 31 - t12 = t1

t1 + d(19) = 83

31 - t12 + d(19) = 83

- t12 + d(19) = 52

t12 = 52 - d(19)

hmm

Answer 4

got a bad negative :)

t12 = -52 + d(19)

Answer 5

should we start at t0?

the mean of t0 and t40 is t20 then

Answer 6

but thats 41 terms ... not 40

Answer 7

sorry i have no idea but thx for trying to help :)

Answer 8

\[2t_1+11d=31\]\[t_1+19d=83\]
\[d = \frac{-2(83)+31}{-2(19)+11}=5\]
\[t_1=83-19(5)=-12\]
\[t_{40}=-12+5(39)=183\]
\[20(-12+183)=3420\]

Answer 9

somethings off just a tad

Answer 10

nah, it should be fine

Answer 11

yeah thats the answer thanks!