2001 CEMC Pascal Problems/Problem 8

Revision as of 10:45, 5 July 2025 by Anabel.disher (talk | contribs) (Created page with "==Problem== In the diagram, the value of <math>x</math> is {{Image needed}} <math> \text{ (A) }\ 100 \qquad\text{ (B) }\ 65 \qquad\text{ (C) }\ 80 \qquad\text{ (D) }\ 70 \qqua...")
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

Problem

In the diagram, the value of $x$ is


An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.


$\text{ (A) }\ 100 \qquad\text{ (B) }\ 65 \qquad\text{ (C) }\ 80 \qquad\text{ (D) }\ 70 \qquad\text{ (E) }\ 50$

Solution 1

We can notice that $5 = 5x^{0}$, allowing us to see that the next term is the previous term but the coefficient is increased by $1$ each time, and the exponent is increased by $1$. This means that the coefficient and exponent must have increased $49$ times from the 1st term to the 50th term.

$5 + 49 = 54$, so the coefficient on the 50th term must be $54$.

$0 + 49 = 49$, so the exponent on the 50th term must be $49$.

This corresponds with $\boxed {\textbf {(A) } 54x^{49}}$.

~anabel.disher

Solution 2

We can use the information found in solution 1 to find the expression for the $n$th term of the sequence. Let $a_n$ be the $n$th term of the sequence

Since the coefficient is increased by $1$ after each term, the coefficient must be $n + c$, where $c$ is some number. Using the first term of the sequence, we can see that $1 + c = 5$, so $c = 4$.

Since the exponent is increased by $1$ after each term, the exponent must be $n + d$, where $d$ is some number. Using the first term of the sequence, we can see that $1 + d = 0$, so $d = -1$.

This means that the formula for the $n$th term is:

$a_n = (n + 4) \times x^{n - 1}$

Plugging in $n = 50$, we get:

$a_{50} = (50 + 4) \times x^{50 - 1} = \boxed {\textbf {(A) } 54x^{49}}$

~anabel.disher