Rakesh and Tessa were asked to find an explicit formula for the sequence $100, 50, 25, 12.5,\ldots$ , where the first term should be $f(1)$ . Rakesh said the formula is $f(n)=100\cdot\left(\dfrac{1}{2}\right)^{n-1}$ , and Tessa said the formula is $f(n)=200\cdot\left(\dfrac{1}{2}\right)^{n}$ . Which one of them is right?

View step-by-step help

Home

Math Problems

Grade 8

Add and subtract rational numbers: word problems

Full solution

Q. Rakesh and Tessa were asked to find an explicit formula for the sequence

100, 50, 25, 12.5,\ldots

, where the first term should be

f(1)

. Rakesh said the formula is

f(n)=100\cdot\left(\dfrac{1}{2}\right)^{n-1}

, and Tessa said the formula is

f(n)=200\cdot\left(\dfrac{1}{2}\right)^{n}

. Which one of them is right?

Check Rakesh's Formula: To determine which formula is correct, we need to check if each formula correctly calculates the terms of the sequence when $n$ is substituted with the term number.
Test Rakesh's Formula: Let's test Rakesh's formula: $f(n) = 100 \times (\frac{1}{2})^{(n-1)}$ . For the first term where $n=1$ , $f(1)$ should equal $100$ . $\newline$ f(\(1) = $100$ \times (\frac{ $1$ }{ $2$ })^{( $1$ $-1$ )} = $100$ \times (\frac{ $1$ }{ $2$ })^ $0$ = $100$ \times $1$ = $100$ \. $\newline$ This matches the first term of the sequence.
Test Rakesh's Formula: Now let's test the second term with Rakesh's formula where $n=2$ . The second term should be $50$ .f(\(2) = $100$ \times ( $1$ / $2$ )^{ $2$ $-1$ } = $100$ \times ( $1$ / $2$ )^ $1$ = $100$ \times $1$ / $2$ = $50$ \.This matches the second term of the sequence.
Test Tessa's Formula: We will continue to test Rakesh's formula for the third term where $n=3$ . The third term should be $25$ . $f(3) = 100 \times (1/2)^{3-1} = 100 \times (1/2)^2 = 100 \times 1/4 = 25$ .This matches the third term of the sequence.
Test Tessa's Formula: Now let's test Tessa's formula: $f(n) = 200 \times (\frac{1}{2})^n$ . For the first term where $n=1$ , $f(1)$ should equal $100$ . $\newline$ $f(1) = 200 \times (\frac{1}{2})^1 = 200 \times \frac{1}{2} = 100.$ $\newline$ This also matches the first term of the sequence.
Test Tessa's Formula: Let's test the second term with Tessa's formula where $n=2$ . The second term should be $50$ .f(\(2) = $200$ \times ( $1$ / $2$ )^ $2$ = $200$ \times $1$ / $4$ = $50$ \. This matches the second term of the sequence.
Evaluate Formulas: We will continue to test Tessa's formula for the third term where $n=3$ . The third term should be $25$ . $f(3) = 200 \times (1/2)^3 = 200 \times 1/8 = 25$ .This matches the third term of the sequence.
Adjust for Sequence: Both formulas have produced the correct first three terms of the sequence. However, we need to check if they will continue to produce the correct terms for all $n$ . The sequence is halving each term, so the base of the exponent should be $\frac{1}{2}$ . The exponent should decrease by $1$ each time to account for the sequence starting at the first term, not the zeroth term.
Adjust for Sequence: Both formulas have produced the correct first three terms of the sequence. However, we need to check if they will continue to produce the correct terms for all $n$ . The sequence is halving each term, so the base of the exponent should be $1/2$ . The exponent should decrease by $1$ each time to account for the sequence starting at the first term, not the zeroth term.Rakesh's formula has the exponent $(n-1)$ , which correctly adjusts for the sequence starting at $n=1$ . Tessa's formula does not have this adjustment, so it will not produce the correct terms for all $n$ .