A linear sequence has a common difference of 8.Three consecutive terms in the sequence are added together to give a total of 126 find the first three terms

34, 42, 50

Step-by-step explanation:

Let the first term be a

Second term will be a+d

Third term will be a+2d

a+a+d+a+2d= 126

3a+3d = 126

Recall that the common difference (d) is 8.

3a+3d= 126

3a+(3×8) = 126

3a+24 = 126

3a= 126 - 24

3a = 102

a = 102/3

a= 34

First term is 34

Second term will be a+d

= 34+8 = 42

Third term will be a+2d

= 34 + (2×8)

= 34+16

= 50

answer: 1,380

step-by-step explanation:

1: you turn %15 into a decimal by moving the decimal point two places to the left.

2: and then you multiply .15 by 1,200 , which will get you 180.00

3: from there you will add 180 to 1,200

4: which will get you 1,380

hope it *im kind of slow so idk if my answer is right*