In each of the​ following, list three terms that continue the arithmetic or geometric sequences. Identify the sequences as arithmetic or geometric. a. 2, 6, 18, 54, 162 b. 1, 8 ,15, 22, 29 c. 11, 15, 19, 23, 27

Step 1 a. Given sequence is 2, 6, 18, 54, 162 Since ratio of two consecutive terms is constant , therefore, sequence is geometric and common ratio is $r=6/2$ , that is $r=3$ Therefore, next term can be obtained by multiplying the previous term by r. Therefore, next three terms are $162\times 3=486$
$486\times 3=1458$
$1458\times 3=4374$ Step 2 b. Given sequence is 1, 8 ,15, 22, 29 Since difference of two consecutive terms is constant , therefore, sequence is arithmetic and common difference is $d=8-1=7,$ Therefore, next term can be obtained by adding the previous term by d. Therefore, next three terms are $29+7=36$ 
$36+7=43,$
$43+7=50$ Step 3 c. Given sequence is 11, 15, 19, 23, 27 Since difference of two consecutive terms is constant , therefore, sequence is arithmetic and common difference is $d=15-11=4,$ Therefore, next term can be obtained by adding the previous term by d. Therefore, next three terms are $27+4=31$
$31+4=35$
$35+4=39$