Find First Term and Common Ratio of Geometric Progression | Step-by-Step Guide

[gunblaze]

ok, the qn goes like this..

Find the first term and common ration of a geometric progression if the sum to n is given by
6 - 2/(3^(n-1))

I tried solving by making both terms of the eqn having the same denominator by multiplying (3^(n-1)) to the first term 6 and then by taking out the 2, i am able to make the formula to 2(3^n - 1)/ (3^n-1). But what about the denominator, i remember that there must be no n at the bottom. Only r - 1 where in this case, r is to be found..

Anyone out here can give me some guide? Thanks for any help given.

 

[HallsofIvy]

First, you have written that wrong. The denominator is not 3^n- 1, it is 3^(n-1). You should have
$$\frac{6(3^{n-1})- 2}{3^{n-1}}= 2(3^n- 1}{3^{n-1}$$

But it's really not necessary to use a general formula. Since you only need to find two numbers, a and r for the general term arⁿ, you really only need two equations. The first term, with n= 1, is a. The formula you are given says it must be
$$a= 6- \frac{2}{3^{1-1}}$$
The sum of the first two terms, with n= 2 is a+ ar and so
$$a+ ar= 6- \frac{2}{3^{2-1}}$$
Can you solve those two equations for a and r?

 

[Architeuthis Dux]

I would approach this problem by first understanding what a geometric progression is. A geometric progression is a sequence of numbers where each term is obtained by multiplying the previous term by a constant number, known as the common ratio. In this case, we can see that the given equation is in the form of a geometric progression, with the first term being 6 and the common ratio being 1/3.

To find the common ratio, we can use the formula for the sum of a geometric progression, which is Sn = a(1-r^n)/(1-r), where Sn is the sum of the first n terms, a is the first term, and r is the common ratio. In this case, we know that Sn is equal to 6 - 2/(3^(n-1)), so we can substitute this into the formula and solve for r.

6 - 2/(3^(n-1)) = a(1-r^n)/(1-r)

We can then solve for r by multiplying both sides by (1-r) and simplifying:

6(1-r) - 2/(3^(n-1))(1-r) = a(1-r^n)
6 - 6r - 2/(3^(n-1)) + 2r/(3^(n-1)) = a - ar^n

We can then rearrange the equation to solve for r:

6 - a = r(6 - 2/(3^(n-1)))

r = (6 - a)/(6 - 2/(3^(n-1)))

We know that a = 6, so we can substitute this in and simplify to get the common ratio:

r = (6 - 6)/(6 - 2/(3^(n-1))) = 0/(6 - 2/(3^(n-1))) = 0

Therefore, the common ratio is 0 and the first term is 6. This means that the geometric progression is simply a sequence of 6's, as the common ratio of 0 means that each term is equal to the previous term.

In conclusion, the first term of the geometric progression is 6 and the common ratio is 0. This means that the sequence is 6, 6, 6, and so on. I hope this explanation helps you understand how to find the first term and common ratio of a geometric progression.