Math Problem Statement

  1. 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?

Solution

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 nn-th term is given by:

an=a1×rn1a_n = a_1 \times r^{n-1}

Where:

  • ana_n is the nn-th term,
  • a1a_1 is the first term,
  • rr is the common ratio,
  • nn is the term number.

Given:

  • a3=8a_3 = 8
  • a6=64a_6 = 64

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

a1×r2=8(1)a_1 \times r^{2} = 8 \quad \text{(1)} a1×r5=64(2)a_1 \times r^{5} = 64 \quad \text{(2)}

To find rr, divide equation (2) by equation (1):

a1×r5a1×r2=648\frac{a_1 \times r^{5}}{a_1 \times r^{2}} = \frac{64}{8}

Simplifying this:

r3=8r^{3} = 8

Taking the cube root of both sides:

r=2r = 2

Now that we know r=2r = 2, substitute this back into equation (1) to find a1a_1:

a1×22=8a_1 \times 2^{2} = 8 a1×4=8a_1 \times 4 = 8 a1=2a_1 = 2

Thus, the first term a1=2a_1 = 2 and the common ratio r=2r = 2.

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

The sum SnS_n of the first nn terms in a geometric sequence is given by:

Sn=a1×(rn1)r1S_n = \frac{a_1 \times (r^n - 1)}{r - 1}

For n=9n = 9:

S9=2×(291)21S_9 = \frac{2 \times (2^9 - 1)}{2 - 1} S9=2×(5121)S_9 = 2 \times (512 - 1) S9=2×511=1022S_9 = 2 \times 511 = 1022

So, the sum of the first 9 terms is 10221022.

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

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

S22=a1×(r221)r1S_{22} = \frac{a_1 \times (r^{22} - 1)}{r - 1} S22=2×(2221)21S_{22} = \frac{2 \times (2^{22} - 1)}{2 - 1} S22=2×(2221)S_{22} = 2 \times (2^{22} - 1)

Now, calculate 2222^{22}:

222=41943042^{22} = 4194304

So,

S22=2×(41943041)=2×4194303=8388606S_{22} = 2 \times (4194304 - 1) = 2 \times 4194303 = 8388606

Thus, the sum of the first 22 terms is 8,388,6068,388,606.


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.

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Math Problem Analysis

Mathematical Concepts

Geometric Sequences
Common Ratio

Formulas

Sum of geometric series

Theorems

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Suitable Grade Level

High School