Problem 60 - Precalculus Mathematics for Calculus

Question

An arithmetic sequence has first term \(a_{1}=1\) and fourth term \(a_{4}=16 .\) How many terms of this sequence must be added to get \(2356 ?\)

Step-by-Step Solution

Answer

You need to add 32 terms to get 2356.

Step 1: Identify the common difference

The formula for the nth term of an arithmetic sequence is \( a_n = a_1 + (n-1) \cdot d \), where \( a_1 \) is the first term and \( d \) is the common difference. We know \( a_4 = a_1 + 3d \). Given \( a_1 = 1 \) and \( a_4 = 16 \), we can set up the equation: \( 1 + 3d = 16 \). Solving for \( d \), we find: \( 3d = 15 \) and consequently \( d = 5 \).

Step 2: Find the nth term with a partial sum formula

The sum of the first \( n \) terms of an arithmetic sequence is given by \( S_n = \frac{n}{2} \cdot (2a_1 + (n-1)d) \). We know \( S_n = 2356 \). Substituting \( a_1 = 1 \) and \( d = 5 \) into the formula, we have \( 2356 = \frac{n}{2} \cdot (2 \cdot 1 + (n-1) \cdot 5 ) \). Simplifying, \( 2356 = \frac{n}{2} \cdot (2 + 5n - 5) \) leads to \( 2356 = \frac{n}{2} \cdot (5n - 3) \).

Step 3: Derive a quadratic equation

From the result of Step 2: \( 2356 = \frac{n}{2} \cdot (5n - 3) \), we multiply both sides by 2 to eliminate the fraction: \( 4712 = n(5n - 3) \), simplifying to \( 4712 = 5n^2 - 3n \). This is a quadratic equation in the form \( 5n^2 - 3n - 4712 = 0 \).

Step 4: Solve the quadratic equation

To find \( n \), solve the quadratic equation \( 5n^2 - 3n - 4712 = 0 \) using the quadratic formula: \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 5 \), \( b = -3 \), and \( c = -4712 \). Calculate the discriminant: \( b^2 - 4ac = (-3)^2 - 4\cdot5\cdot(-4712) \). After calculation, the discriminant results in a positive value, confirming real solutions. Substitute into the quadratic formula to find \( n \) values.

Step 5: Verify the positive solution for n

Using the positive solution from the quadratic formula, we find \( n = 32 \). Verify by back-substitution, checking calculations, and ensuring that a non-decimal, positive solution is appropriate for the context of adding terms of a sequence.

Key Concepts

Common Difference

In an arithmetic sequence, the common difference is a key characteristic. This is the amount that each term increases by as you move from one term to the next. It's like a steady step up the number line. To find this in any given sequence, you subtract the first term from the second term, the second from the third, and so on. For instance, in the problem where the first term is 1 and the fourth term is 16, the common difference can be calculated as follows:

• Formula setup: We use the formula for the nth term: \( a_n = a_1 + (n-1) \cdot d \).
• Plug in known values: Given \( a_1 = 1 \) and \( a_4 = 16 \), we have: \( 1 + 3d = 16 \).
• Solve for \( d \): Simplify to find: \( 3d = 15 \), hence \( d = 5 \).

The sequence increases by 5 each step. That's the common difference.

Nth Term Formula

Calculating specific terms in an arithmetic sequence is made easy with the nth term formula. This formula lets you find any term's value knowing its position (n) in the sequence. It is expressed as \( a_n = a_1 + (n-1) \cdot d \).

• First, identify your first term (\( a_1 \)). In the provided example, \( a_1 = 1 \).
• Next, use the common difference (\( d \)) you found earlier; here, it is 5.
• Finally, plug these values into the nth term formula to find any desired term. For example, \( a_4 \) would be calculated as \( 1 + 3 \cdot 5 = 16 \).

With this formula, finding any term in the sequence becomes a breeze.

Sum of Arithmetic Sequence

To find how many terms we need to sum in a sequence, we use the arithmetic sequence sum formula, which is \( S_n = \frac{n}{2} \cdot (2a_1 + (n-1)d) \). This allows finding the sum of the first \( n \) terms.

• We know \( S_n = 2356 \), the desired sum, and solve for \( n \).
• The formula becomes:\( 2356 = \frac{n}{2} \cdot (2 \cdot 1 + (n-1) \cdot 5) \).
• Simplified, this gives \( 2356 = \frac{n}{2} \cdot (5n - 3) \).

By working through the algebra, you can establish an equation to solve for \( n \). This is how you determine how many terms must be added to yield a specific total.

Quadratic Equation

Quadratic equations are used to find number solutions in problems involving polynomials of degree 2. The general form is \( ax^2 + bx + c = 0 \). In our scenario, once the sum formula was simplified, it led us to \( 5n^2 - 3n - 4712 = 0 \).

• Identify the coefficients: Here, \( a = 5 \), \( b = -3 \), and \( c = -4712 \).
• Use the quadratic formula: \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
• Calculate the discriminant: \( b^2 - 4ac = (-3)^2 - 4 \cdot 5 \cdot (-4712) \), which is positive, ensuring real solutions exist.
• After plugging into the formula, solve for \( n \) to find usable values in the context provided.

Using these steps, we find \( n \) which fits the conditions of the problem, ensuring the arithmetic sequence sums to the desired amount.