Question

7D%5C%5C%20%26%5C%7B10%2C12%2C14.4%2C17.28%2C%20%5Cldots%5C%7D%20%5Ctext%20%7B%20exceed%20%7D%20100%20%3F%20%5Cend%7Baligned%7D" id="TexFormula1" title="\begin{aligned} &\text { 55. At which term does the sequence }\\ &\{10,12,14.4,17.28, \ldots\} \text { exceed } 100 ? \end{aligned}" alt="\begin{aligned} &\text { 55. At which term does the sequence }\\ &\{10,12,14.4,17.28, \ldots\} \text { exceed } 100 ? \end{aligned}" align="absmiddle" class="latex-formula">

Answer 1

Answer:

14th term

Step-by-step explanation:

12÷10 = 1.2

14.4÷12 = 1.2

17.28÷14.4 = 1.2

Then it’s a geometric sequence of first term ‘V₀=10’ and common ration ‘q = 1.2’

We call it (Vn)

The general term of the sequence:

Vn = V₀×qⁿ⁻¹ = 10×(1.2)ⁿ⁻¹

To determine the term at which the sequence exceeds 100 we have to solve the equation:

Vn ≥ 100

⇔ 10×(1.2)ⁿ⁻¹ ≥ 100

⇔ (1.2)ⁿ⁻¹ ≥ 100/10

⇔ (1.2)ⁿ⁻¹ ≥ 10

⇔ ln(1.2)ⁿ⁻¹ ≥ ln(10)

⇔ (n-1)×ln(1.2) ≥ ln(10)

$\Longleftrightarrow n-1\geq\frac{ln10}{ln1.2}$

$\Longleftrightarrow n-1\geq12.6\\\Longleftrightarrow n\geq13.6\\\Longrightarrow n=14$

Verification:

$V_{14}=10\times \left( 1.2\right)^{14-1} =106.993205379072$