What is the sum of a 14–term arithmetic sequence where the last term is 30 and the common difference is –5?
do you know the formula? lol
no i cant tell the difference between the 2 formulas
\[\LARGE S_n=\frac n2[2a_1+(n-1)d]\] you're dealing with this one.... and this one \[\LARGE a_n=a_1+(n-1)d\]
yeah idk how to tell which one applies to which equation.
from the second equation you have to find a_1 ... then to substitute into the first one , to find their sum.
where does the -5 go ?
d=-5
is An = 30 ? or 14
you have to find a_1 ...
is it - 145?
\[\LARGE a_{14}=a_1+(14-1)(-5)\] \[\LARGE 30=a_1+13(-5)\] \[\LARGE 30=a_1-65\] \[\LARGE a_1=30+65\] \[\LARGE a_1=95\]
now \[\LARGE S_{14}=\frac{14}{2}[2\cdot 95+(14-1)\cdot (-5)]\] can you do it? :)
see i dont get why it was 14 first then changed to 30 and no im not good with fractions
it changed to 30 because statement says: What is the sum of a 14–term ...where the last term is 30 so it means a_14=30 that's why a_14 changed to 30
oh okayy, see i didnt know which one went where i didnt know that 14= 30
and don't worry about fractions... \[\LARGE S_{14}=\frac{14}{2}[2\cdot 95+(14-1)\cdot (-5)]\] \[\LARGE S_{14}=7\cdot [2\cdot 95+(14-1)\cdot (-5)]\] now you don't have fractions ;) .. can you give it a try?
875
I don't know... give me a second to calculate it lol
okay
yes... 875 is correct . Well done :)
thank you
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