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Find S20 when t1 = 4 and d = 5
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Now you have to use formula
tn=t1+(n-1)d
\(S_{n}=\frac{n}{2} (2 t_1 + (n-1)d)\) sub n=20, and everything you've got there. do u need further help?
cool
S(20=20/2(2(4)+(20-1)5)
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And how would I work this out @Shadowys
um,simplify this then. \(\frac{20}{2}[(2)(4)+(20-1)5]=10[8+19(5)]\) what follows can be done by manual arithmetic or a calculator.
I got 1030
yup, that's what I got too. :)
Cool
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