How many terms of g. p. 3, 3², 3³ are needed to give the sum 120?
(show your work)! (90 points) !

The Given G.P. is 3, 3², 3³,
Let n terms of this G.P. be required to obtain the sum as 120

Sₓ = a(rˣ - 1) / r-1
Here, a = 3 and r = 3

Substitute the values, into the expression:

Sₓ = 120 = 3(3ˣ - 1) / 3-1
⇒120 =  3(3ˣ - 1) / 2
⇒120×2 / 3 = 3ˣ - 1
⇒3ˣ - 1 = 80
⇒3ˣ = 81
⇒3ˣ = 3⁴
So, n = 4

Thus, Four terms of the given G.P. are required to obtain the sum as 120
In short, Your Final Answer is 4

Hope this helps!

First term a = 3,

Common ratio = 3² / 3 = 3

Formula for sum of GP, S = Sum.

S = a(rⁿ - 1) / (r - 1).      We are to solve when S = 120

120 = 3(3ⁿ - 1) / (3 - 1)

120 = 3(3ⁿ - 1) /2

3(3ⁿ - 1) /2 = 120

(3ⁿ - 1) =  2*120/3 = 2*40

(3ⁿ - 1) =  80

3ⁿ  =  81

3ⁿ  =  3⁴

Since the bases are the same, hence the indices will be same.

n = 4

Therefore 4 of the terms are needed.

This could also been solved by adding each of the terms, until 120 is gotten

3 = 3

3² = 3*3 = 9

3³ = 3*3*3 = 27

3⁴ = 3*3*3*3 = 81

Sum = 3 + 9 + 27 + 81 = 120

Hope this explains it.