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

Question

amistre64 · Answer

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

amistre64 · Answer

or do we start at a t0?

amistre64 · Answer

186/6 = 31 - t12 = t1

t1 + d(19) = 83

31 - t12 + d(19) = 83

- t12 + d(19) = 52

t12  = 52 - d(19)

hmm

amistre64 · Answer

got a bad negative :)

t12 = -52 + d(19)

amistre64 · Answer

should we start at t0?

the mean of t0 and t40 is t20 then

amistre64 · Answer

but thats 41 terms ... not 40

anonymous · Answer

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

amistre64 · Answer

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

amistre64 · Answer

somethings off just a tad

amistre64 · Answer

nah, it should be fine

anonymous · Answer

yeah thats the answer thanks!