The sum of the first three terms of a geometric sequence is . The sum of the
fourth, fifth and sixth term of the same sequence is . Determine the value of
common ratio .

Ask by Pollard Garza. in South Africa

Mar 21,2025

Question

The sum of the first three terms of a geometric sequence is . The sum of the
fourth, fifth and sixth term of the same sequence is . Determine the value of
common ratio .

Ask by Pollard Garza. in South Africa

Mar 21,2025

UpStudy AI · Accepted Answer

We start by letting the first term of the geometric sequence be $$ a $$ and the common ratio be $$ r $$. Then, the first three terms are:

$$
a,\quad ar,\quad ar^2,
$$

and their sum is

$$
a(1+r+r^2)= \frac{63}{2}.
$$

Similarly, the fourth, fifth, and sixth terms are:

$$
ar^3,\quad ar^4,\quad ar^5,
$$

and their sum is

$$
ar^3(1+r+r^2)= \frac{63}{16}.
$$

To eliminate the common factor $$ a(1+r+r^2) $$, we divide the second equation by the first:

$$
\frac{ar^3(1+r+r^2)}{a(1+r+r^2)}=\frac{\frac{63}{16}}{\frac{63}{2}}.
$$

Simplifying the left side, we obtain:

$$
r^3 = \frac{63/16}{63/2}.
$$

Evaluate the right side:

$$
r^3 = \frac{63}{16} \div \frac{63}{2} = \frac{63}{16} \times \frac{2}{63} = \frac{2}{16} = \frac{1}{8}.
$$

Now, solving for $$ r $$, we have:

$$
r^3=\frac{1}{8} \quad \Longrightarrow \quad r=\frac{1}{2}.
$$

Thus, the common ratio is

$$
\boxed{\frac{1}{2}}.
$$
The common ratio $$ r $$ is $$ \frac{1}{2} $$.