01. The third term of a multiple is 8 and the sixth term is 64. Here

(a) Find the first term and the common ratio.

(b) Find the sum of the original 9 terms.

(๑) What is the sum of 22 when the terms are added?

To solve this problem, we first need to determine whether the sequence is arithmetic or geometric. Given that the third term is 8 and the sixth term is 64, it suggests the sequence is geometric, where each term is obtained by multiplying the previous term by a constant called the **common ratio**.

### (a) Find the first term and the common ratio.

In a geometric sequence, the \( n \)-th term is given by:

\[
a_n = a_1 \times r^{n-1}
\]

Where:
- \( a_n \) is the \( n \)-th term,
- \( a_1 \) is the first term,
- \( r \) is the common ratio,
- \( n \) is the term number.

Given:
- \( a_3 = 8 \)
- \( a_6 = 64 \)

We can write the equations for the third and sixth terms:

\[
a_1 \times r^{2} = 8 \quad \text{(1)}
\]
\[
a_1 \times r^{5} = 64 \quad \text{(2)}
\]

To find \( r \), divide equation (2) by equation (1):

\[
\frac{a_1 \times r^{5}}{a_1 \times r^{2}} = \frac{64}{8}
\]

Simplifying this:

\[
r^{3} = 8
\]

Taking the cube root of both sides:

\[
r = 2
\]

Now that we know \( r = 2 \), substitute this back into equation (1) to find \( a_1 \):

\[
a_1 \times 2^{2} = 8
\]
\[
a_1 \times 4 = 8
\]
\[
a_1 = 2
\]

Thus, the first term \( a_1 = 2 \) and the common ratio \( r = 2 \).

### (b) Find the sum of the original 9 terms.

The sum \( S_n \) of the first \( n \) terms in a geometric sequence is given by:

\[
S_n = \frac{a_1 \times (r^n - 1)}{r - 1}
\]

For \( n = 9 \):

\[
S_9 = \frac{2 \times (2^9 - 1)}{2 - 1}
\]
\[
S_9 = 2 \times (512 - 1)
\]
\[
S_9 = 2 \times 511 = 1022
\]

So, the sum of the first 9 terms is \( 1022 \).

### (๑) What is the sum of 22 when the terms are added?

Here, we need to find the sum of the first 22 terms:

\[
S_{22} = \frac{a_1 \times (r^{22} - 1)}{r - 1}
\]
\[
S_{22} = \frac{2 \times (2^{22} - 1)}{2 - 1}
\]
\[
S_{22} = 2 \times (2^{22} - 1)
\]

Now, calculate \( 2^{22} \):

\[
2^{22} = 4194304
\]

So,

\[
S_{22} = 2 \times (4194304 - 1) = 2 \times 4194303 = 8388606
\]

Thus, the sum of the first 22 terms is \( 8,388,606 \).

---

Would you like further details or have any questions?

Here are some related questions:

1. How does the sum formula for a geometric series change if the common ratio is less than 1?
2. What is the sum of the first 15 terms for the same sequence?
3. How do you find the sum of an infinite geometric series?
4. How would you solve the problem if the sequence were arithmetic instead of geometric?
5. What is the tenth term of the sequence?

**Tip:** Understanding the behavior of geometric sequences is crucial, especially how the common ratio affects the growth or decay of the sequence.

Learn how to solve geometric sequence problems, including finding the first term, common ratio, and sum of terms. This detailed solution covers specific examples where the third term is 8 and the sixth term is 64, demonstrating step-by-step calculations. Suitable for high school students.

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