help please
find the indicated sum for the arithmetic series
problem below ...

Question

anonymous · Answer

$$\sum_{k=1}^{12} (-2+6k)$$

anonymous · Answer

a) 444
b) 368
c) 888
d)565

campbell_st · Answer

find the 1st term when k = 1 and the last term when k = 12

1st term   a = -2+6 * (1) = 4

last term  l =  -2 + 6*(12) = 70

the sum of the terms is 

$$S_{n} = \frac{n}{2}(a + l)$$

you found a and l above, and n = 12 

substitute and evaluate.

anonymous · Answer

its 444

anonymous · Answer

how'd you get that @breeza

anonymous · Answer

where would you substitute the number ? @campbell_st

anonymous · Answer

because on that bottom equation, 
he's telling you n=12 
and
a= the first term when u substitute 1 so 
a=4
and
I= the last term you substituted and they wan you 2 substitute 12 so
I=70

anonymous · Answer

juist plug in the numbers he gave you

campbell_st · Answer

if you read the information and plug in the values its 

$$S_{12} = \frac{12}{2}[ 4 + 70]$$