Question

How do you find the sum of the first four terms of the G.P.

Answer 1

We want to calculate the sum of the first four terms of the following G.P.

`\sum ^(\square)_13^(n-1)`

There are two ways of solving this question. The first one would be to brute force calculate the first four terms and sum them(since it is a simple general term, it is totally possible to do it), or use the formula for the sum of the first n terms.

We can do both to be certain of our result.

Doing by brute force, we have the first 4 terms as:

`\begin{gathered} a_1=3^(1-1)=3^0=1 \\ a_2=3^(2-1)=3^1=3 \\ a_3=3^(3-1)=3^2=9 \\ a_4=3^(4-1)=3^3=27 \\ a_1+a_2+a_3+a_4=1+3+9+27=40 \end{gathered}`

The general formula to sum the first n terms is

`S_n=(a_1(r^n-1))/(r-1)`

Where n represents the amount of terms we want to sum, a_1 is the first term, and r is the base of our geometric progression.

For our problem, we have an n = 4, a_1 = 1 and r = 3. Plugging those values in our formula, we have

`S_4=(1*(3^4-1))/(3-1)=(80)/(2)=40_{}`

The sum of the first four terms of this progression is equal to 40.