Write the first six terms of the geometric sequence with the first term 6 and common ratio 1/3

6, 2, 2/3, 2/9, 2/27, 2/81

Step-by-step explanation:

The nth term of a geometric progression is expressed as;

Tn  = ar^n-1

a is the first term

n is the number of terms

r is the common ratio

Given

a = 6

r = 1/3

when n = 1

T1 = 6(1/3)^1-1

T1 = 6(1/3)^0

T1 = 6

when n = 2

T2= 6(1/3)^2-1

T2= 6(1/3)^1

T2 = 2

when n = 3

T3 = 6(1/3)^3-1

T3= 6(1/3)^2

T3= 6 * 1/9

T3 = 2/3

when n = 4

T4 = 6(1/3)^4-1

T4= 6(1/3)^3

T4= 6 * 1/27

T4 = 2/9

when n = 5

T5 = 6(1/3)^5-1

T5= 6(1/3)^4

T5= 6 * 1/81

T5 = 2/27

when n = 6

T6 = 6(1/3)^6-1

T6= 6(1/3)^5

T6= 6 * 1/243

T6 = 2/81

Hence the first six terms are 6, 2, 2/3, 2/9, 2/27, 2/81

idk u tell me

step-by-step explanation:

the answer is 40.

proof: