1)find the sum of the arithmetic sequence. -10, -7, -4, -1, 2, 5, 8 a)11 b)-70 c)0 d)-7 2)find the sum of the geometric sequence. 1, 1/4, 1/16,1/64,1/256 a)341 b)1/192 c)1/768 d)341/256 3)write the sum using summation notation, assuming the suggested pattern continues. -9 - 4 + 1 + 6 + + 66 4)write the sum using summation notation, assuming the suggested pattern continues. 25 + 36 + 49 + 64 + + n2 + 5)write the sum using summation notation, assuming the suggested pattern continues. 8 - 40 + 200 - 1000 + .. 6)find the sum of the first 12 terms of the sequence. show all work for full credit. 1, -4, -9, -14, . .

See below

Step-by-step explanation:

1)  Sum = (n/2){a1 + L]     where  n =  number count, a1 = first term and L = last term

so here its is (7/2) [ -10 + 8]

= 7/2 * -2

= -7  (answer)

2/

Sum of n terms  =   a1 * (1 - r^n)/ (1 - r)   where r = common ratio

Here r =  1/4

so its   1 * ( 1 - 1/4^5) / 1 - 1/4

= 341 / 256

3.

This is an arithmetic sequence  with first term -9,  last term 66  and common difference 5.

16

∑ (-9 + 5(n - 1)

n=1

Note the 16 comes from  66 being the 16th term  ( solve 66 = -9 + 5(n - 1)

5  This is geometric with common ratio -5 , first term 8  which continues without bounds.

∞

∑ 8(-5)^(n-1)

n=1

6. This is arithmetic with a1 = 1 and d = -5

Sn  = (n/2) [ 2a1 + (n - 1)d]

So S12 = 6( 2 + 11*-5)

= 6 *  -53

= -318 ( answer)

1. is 4 and 2. is 3

step-by-step explanation:

so if you look at the first one you see that 1 is to 4 and 2 is to 8 and so on, so you can see for every 1 there is a times 4 so if it were 3 then it would be 3*4 and it is 12

for number 2 it is 3 because 1 is to 3 and 2 is to 6 so for every 1 there is 3 so if it were 3 then there would be 3*3 = 9