Math Problem Statement

Use the nth term of a geometric sequence An=a1(r n-1) to answer the following questions 1.What is the 5th term of the geometric sequence 3/20,3/2,15....? 2.Find the six term of a geometric sequence where the second term is 6 and the common ratio is 2. 3.Find K so that terms k-3,k+1 and 4k-2 form a geometric sequence. 4.In the geometric sequence 6,12,24,48,...., Which term is 768? 5.The second term of a geometric is 5/4 and it's fourth term is

Solution

Let's solve each question step by step using the nth term of a geometric sequence formula:

An=a1r(n1)A_n = a_1 \cdot r^{(n-1)}

where AnA_n is the nth term, a1a_1 is the first term, and rr is the common ratio.

1. What is the 5th term of the geometric sequence 3/20, 3/2, 15...?

Step 1: Find the common ratio rr.

The common ratio rr is found by dividing the second term by the first term:

r=32320=32×203=10r = \frac{\frac{3}{2}}{\frac{3}{20}} = \frac{3}{2} \times \frac{20}{3} = 10

Step 2: Use the nth term formula to find the 5th term.

Here, a1=320a_1 = \frac{3}{20}, r=10r = 10, and n=5n = 5.

A5=320×10(51)=320×104=320×10000=1500A_5 = \frac{3}{20} \times 10^{(5-1)} = \frac{3}{20} \times 10^4 = \frac{3}{20} \times 10000 = 1500

So, the 5th term is 1500.

2. Find the 6th term of a geometric sequence where the second term is 6 and the common ratio is 2.

Step 1: Find the first term a1a_1.

Given the second term (A2A_2) is 6:

A2=a1r(21)=a12=6    a1=3A_2 = a_1 \cdot r^{(2-1)} = a_1 \cdot 2 = 6 \implies a_1 = 3

Step 2: Use the nth term formula to find the 6th term.

Now, a1=3a_1 = 3, r=2r = 2, and n=6n = 6.

A6=3×2(61)=3×25=3×32=96A_6 = 3 \times 2^{(6-1)} = 3 \times 2^5 = 3 \times 32 = 96

So, the 6th term is 96.

3. Find kk so that the terms k3k-3, k+1k+1, and 4k24k-2 form a geometric sequence.

For these terms to form a geometric sequence, the ratio of consecutive terms must be equal:

k+1k3=4k2k+1\frac{k+1}{k-3} = \frac{4k-2}{k+1}

Cross-multiply to solve:

(k+1)2=(k3)(4k2)(k+1)^2 = (k-3)(4k-2)

Expand both sides:

k2+2k+1=4k214k+6k^2 + 2k + 1 = 4k^2 - 14k + 6

Simplify:

0=3k216k+50 = 3k^2 - 16k + 5

Solve this quadratic equation using the quadratic formula:

k=(16)±(16)24×3×52×3=16±256606=16±1966=16±146k = \frac{-(-16) \pm \sqrt{(-16)^2 - 4 \times 3 \times 5}}{2 \times 3} = \frac{16 \pm \sqrt{256 - 60}}{6} = \frac{16 \pm \sqrt{196}}{6} = \frac{16 \pm 14}{6}

k=306=5ork=26=13k = \frac{30}{6} = 5 \quad \text{or} \quad k = \frac{2}{6} = \frac{1}{3}

So, k=5k = 5 or k=13k = \frac{1}{3}.

4. In the geometric sequence 6, 12, 24, 48,...., which term is 768?

Step 1: Find the common ratio rr.

r=126=2r = \frac{12}{6} = 2

Step 2: Use the nth term formula to find nn.

An=62(n1)=768A_n = 6 \cdot 2^{(n-1)} = 768

Solve for nn:

2(n1)=7686=128    2(n1)=27    n1=7    n=82^{(n-1)} = \frac{768}{6} = 128 \implies 2^{(n-1)} = 2^7 \implies n - 1 = 7 \implies n = 8

So, the 8th term is 768.

5. The second term of a geometric sequence is 54\frac{5}{4} and the fourth term is...

Let’s denote the first term as a1=aa_1 = a, the common ratio as rr.

Given:

A2=ar=54A_2 = a \cdot r = \frac{5}{4}

A4=ar3A_4 = a \cdot r^3

To proceed, please provide the fourth term to find the rest of the values.

Would you like further details or have any questions?

Here are some related questions you might consider:

  1. What is the 7th term of a geometric sequence given the first term and common ratio?
  2. How can we find the sum of the first n terms of a geometric sequence?
  3. How does the common ratio affect the growth of a geometric sequence?
  4. What happens if the common ratio is a fraction or negative?
  5. How to determine if a sequence is geometric given a set of terms?

Tip: In a geometric sequence, each term is obtained by multiplying the previous term by a constant (the common ratio). Understanding the common ratio helps to quickly identify the pattern of the sequence!

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Geometric Sequences

Formulas

nth term of a geometric sequence

Theorems

-

Suitable Grade Level

Grades 9-12