Question

Write the first five terms of the geometric sequence if the nth term is given by 36(1/3)^n-1

Answer 1

36, 12, 4, 4/3, 4/9.

Explanation:

We find each term by substituting the sequence number.

So the first term (where n = 1)

= 36(1/3)^(1-1)

= 36(1/3)^0

= 36 * 1

= 36.

The second term is 36(1/3)^(2-1) = 36 * 1/3

= 12.

We can find subsequent terms by just multiplying the previous term by 1/3 so the third term = 12 * 1/3 = 4.

So the first 5 terms are 36, 12, 4, 4/3, 4/9.

Answer 2

G1=36,G2=12,G3=4,G4=4/3,G5=4/9

Explanation:

Since the nth term is given by;

Gn= 36(1/3)^n-1, then, substitute the values of n as 1,2,3,4 and 5 to get the values of G1,G2,G3,G4 and G5 respectively.

G1=36(1/3)^1-1 = 36

G2= 36(1/3)^2-1= 12

G3= 36(1/3)^3-1 = 4

G4= 36(1/3)^4-1 = 4/3

G5= 36(1/3)^5-1=4/9