If the nth partial sum of the sequence an is given by
n
sigma 2k+4
k=1
then what is the nth term of the sequence?
a. 6
b. 2n+4
c. 6+8+10+12++(2n+4)
d. 2(1)+4+2(n)+4

This is an example of a series in arithmetic progression.

The nth term an can be calculated using the formula:

an = a1 + (n – 1) d

where: a1 = is the 1st term of the series

d = common difference

Calculating for the 1st term:

a1 = 2k + 4 = 2 (1) + 4 = 6

Calculating for the common difference by :

d = a2 – a1

d = 2 (2) + 4 – 6

d = 2

Therefore calculating for an:

an = 6 + (n – 1) 2

an = 6 + 2n -2

an = 2n + 4

Therefore the answer is B. 2n+4

cosine

step-by-step explanation:

step-by-step explanation:

i think something is missing in your question because y-1=(x-12) would not pass through the point (-4,-3). because if you replace x by -4 your are supposed to find y=-3 which is not the case her. if you replace x by -4 you find =-15.

i assume the right equation is y-1=1/4*(x-12) because this one is passing through both your points (if you are interested to know how i found that out i can tell you).

in this equation if i replace x by -4 i find y=-3 and if i replace x by 12, i find y=1 so the line representing this equation is passing through those 2 points.

then the standard form of the equation is finding an expression of y in function of x

y-1=1/4*(x-12)

y=x/4-3+1

y=x/4-2

hope that .

do you confirm there was a mistake in the question?