When the same constant is added to the numbers $60,$ $120,$ and $140,$ a three-term geometric sequence arises. What is the common ratio of the resulting sequence?

omgitsrne Feb 22, 2024

#1**+1 **

In general, in a geometric series, three terms, say \(a{r}^{k}, a{r}^{k+1}, a{r}^{k+2}\), then \(a{r}^{k}*a{r}^{k+2}=({a{r}^{k+1}})^{2}\), or, otherwise, in three terms of a geometric series, the middle term squared, equals the product of the two next to it.

Using this, we get the equation

\((60+x)(140+x)={(120+x)}^{2}\), where x is the added number.

\(8400+200x+{x}^{2}=14400+240x+{x}^{2}\). Cancelling out the x^{2}:

\(8400+200x=14400+240x\)

\(40x=-6000\)

\(x=-150\).

The resulting series will be:

\(-90, -30, -10\).

The common ratio is \(\frac{-30}{-90}=\frac{1}{3}\). __So, the resulting sequence has a common ratio of 1/3__.

hairyberry Feb 22, 2024