Question

the relationship between womens whole - number shoe sizes and foot lengths is an arithmetic sequence, where (a_{n}) is the foot length in inches that corresponds to a shoe size of (n). a womens size 7 fits a foot (9\frac{1}{3}) inches long, and a womens size 12 fits a foot 11 inches long. what is the explicit formula for the arithmetic sequence? round to the nearest hundredth, if necessary.
(\bigcirc) (a_{n}approx - 17.91 + 2.98(n - 1))
(\bigcirc) (a_{n}approx - 5 + 2.99(n - 1))
(\bigcirc) (a_{n}approx 7 + 0.33(n - 1))
(\bigcirc) (a_{n}approx 7.35 + 0.33(n - 1))

Explanation:

Step1: Recall Arithmetic Sequence Formula

The explicit formula for an arithmetic sequence is \( a_n = a_1 + d(n - 1) \), where \( a_1 \) is the first term and \( d \) is the common difference. We know two terms: when \( n = 7 \), \( a_7 = 9\frac{1}{3}=\frac{28}{3}\approx9.33 \) inches, and when \( n = 12 \), \( a_{12}=11 \) inches.

Step2: Calculate the Common Difference \( d \)

The number of terms between \( n = 7 \) and \( n = 12 \) is \( 12 - 7 = 5 \). The difference in the terms is \( 11-\frac{28}{3}=\frac{33 - 28}{3}=\frac{5}{3}\approx1.67 \)? Wait, no, wait. Wait, the formula is \( a_n=a_1 + d(n - 1) \), so for \( n = 7 \), \( a_7=a_1 + 6d \) (since \( n-1 = 6 \)), and for \( n = 12 \), \( a_{12}=a_1+11d \). Then subtract the two equations: \( a_{12}-a_7=(a_1 + 11d)-(a_1 + 6d)=5d \). So \( 11-\frac{28}{3}=5d \). \( 11=\frac{33}{3} \), so \( \frac{33}{3}-\frac{28}{3}=\frac{5}{3}=5d \), so \( d=\frac{5}{3}\div5=\frac{1}{3}\approx0.33 \).

Step3: Find \( a_1 \)

Now use \( a_7=a_1 + 6d \). We know \( a_7=\frac{28}{3}\approx9.33 \) and \( d=\frac{1}{3}\approx0.33 \). So \( \frac{28}{3}=a_1+6\times\frac{1}{3} \). \( 6\times\frac{1}{3}=2 \), so \( a_1=\frac{28}{3}-2=\frac{28 - 6}{3}=\frac{22}{3}\approx7.33 \approx7.35 \) (rounded to nearest hundredth).

Step4: Write the Explicit Formula

Now, with \( a_1\approx7.35 \) and \( d\approx0.33 \), the explicit formula is \( a_n = 7.35+0.33(n - 1) \). Let's check the options. The fourth option is \( a_n = 7.35+0.33(n - 1) \), which matches our calculation. Let's verify with \( n = 7 \): \( a_7=7.35+0.33(6)=7.35 + 1.98 = 9.33 \), which is \( 9\frac{1}{3}\approx9.33 \). For \( n = 12 \): \( a_{12}=7.35+0.33(11)=7.35 + 3.63 = 10.98\approx11 \) (close enough with rounding). The other options:

• First option: \( a_n=-17.91 + 2.98(n - 1) \), \( d = 2.98 \) is way too big, incorrect.
• Second option: \( d = 2.98 \), incorrect.
• Third option: \( a_1 = 7 \), then \( a_7=7+0.33(6)=7 + 1.98 = 8.98

eq9.33 \), incorrect.

• Fourth option: \( a_1 = 7.35 \), \( d = 0.33 \), which we calculated, so it's correct.

Answer:

\( a_n = 7.35 + 0.33(n - 1) \) (the fourth option)