Problem 17 - Basic Engineering Mathematics

Question

The first, twelfth and last term of an arithmetic progression are \(4,31 \frac{1}{2}\), and \(376 \frac{1}{2}\) respectively. Determine (a) the number of terms in the series, (b) the sum of all the terms and (c) the 80 'th term.

Step-by-Step Solution

Answer

(a) 150 terms, (b) Sum is 28537.5, (c) 80th term is 201.5.

Step 1: Identify the Terms and Common Difference Formula

The first term is \(a = 4\), the twelfth term is \(a_{12} = 31 \frac{1}{2} = \frac{63}{2}\), and the last term is \(l = 376 \frac{1}{2} = \frac{753}{2}\). The formula for the \(n_{th}\) term, \(a_n\), in an arithmetic progression is \(a_n = a + (n-1)d\), where \(d\) is the common difference.

Step 2: Calculate Common Difference Using the Twelfth Term

We know \(a_{12} = a + 11d\). Substitute the values to get \(\frac{63}{2} = 4 + 11d\). Convert 4 into an improper fraction: \(4 = \frac{8}{2}\). Thus, \(\frac{63}{2} = \frac{8}{2} + 11d\). Solving for \(d\), we get \(\frac{63}{2} - \frac{8}{2} = 11d\), which simplifies to \(\frac{55}{2} = 11d\). Therefore, \(d = \frac{5}{2}\).

Step 3: Find the Number of Terms, \(n\)

Using the last term \(a_n = 376\frac{1}{2} = \frac{753}{2}\), we apply the formula: \(\frac{753}{2} = 4 + (n-1) \frac{5}{2}\). Rewrite 4 as \(\frac{8}{2}\) to get: \(\frac{753}{2} = \frac{8}{2} + (n-1)\frac{5}{2}\). Solve for \(n\) by rearranging to \(\frac{753}{2} - \frac{8}{2} = (n-1)\frac{5}{2}\). This results in \(\frac{745}{2} = (n-1)\frac{5}{2}\). Divide both sides by \(\frac{5}{2}\): \(\frac{745}{2} \times \frac{2}{5} = n-1\), so \(n-1 = 149\) and \(n = 150\).

Step 4: Calculate the Sum of the Series

The sum \(S_n\) of an arithmetic series is \((n/2) \times (a + l)\). Substitute the values: \(S_{150} = \frac{150}{2} \times (4 + \frac{753}{2})\). Rewrite 4 as \(\frac{8}{2}\): \(S_{150} = 75 \times \left(\frac{8}{2} + \frac{753}{2}\right)\). Simplify to \(S_{150} = 75 \times \frac{761}{2}\) or \(S_{150} = 75 \times 380.5 = 28537.5\).

Step 5: Determine the 80th Term

Using the nth term formula \(a_n = a + (n-1)d\), determine the 80th term: \(a_{80} = 4 + 79\times\frac{5}{2}\). Compute \(79 \times \frac{5}{2} = \frac{395}{2}\), and convert 4 to \(\frac{8}{2}\) so \(a_{80} = \frac{8}{2} + \frac{395}{2} = \frac{403}{2} = 201.5\).

Key Concepts

Common difference

When dealing with an arithmetic progression, understanding the common difference is key. This difference, identified as \(d\), is the fixed amount added to each term to reach the next term in the sequence. It remains consistent throughout the progression. To find \(d\), you often subtract a term from the next one or use a specific formula like \(d = \frac{a_n - a}{n-1}\). Here, \(a_n\) is the \(n_{th}\) term, and \(a\) is the first term. In our problem about an arithmetic progression, the common difference was calculated using the 12th term and the first term, resulting in \(d = \frac{5}{2}\). This consistent difference plays a crucial role in determining other properties of the sequence.

Sum of arithmetic series

To find the sum of an arithmetic series, use the sum formula: \[ S_n = \frac{n}{2} \times (a + l) \] where \(n\) is the number of terms, \(a\) is the first term, and \(l\) is the last term of the series. This formula effectively adds the first and last terms, multiplies by the number of terms divided by two, capturing the entire series. For example, in the given exercise, this formula helped find the series' sum as 28,537.5 when \(n = 150\), \(a = 4\), and \(l = 376.5\).

• The formula is simple yet powerful.
• It helps summarize the entire arithmetic progression in one efficient step.

Nth term formula

The nth term formula in an arithmetic progression helps determine any term's value. It is \(a_n = a + (n-1)d\), where:

• \(a_n\) is the \(n_{th}\) term
• \(a\) is the first term
• \(d\) is the common difference

This formula is pivotal for checking specific terms without listing the whole series. For instance, to find the 80th term in the problem, we used the first term \(a = 4\) and the common difference \(d = \frac{5}{2}\). Thus, the 80th term is calculated as 201.5. This allows us to calculate any term in the progression quickly.

Number of terms

The number of terms in an arithmetic progression tells us how many elements are present in the sequence. It's often found using the last term formula rearranged to solve for \(n\), the number of terms. For instance, \(a_n = a + (n-1)d\) can be rewritten to find \(n\) when the last term, \(a_n\), is known. In the given problem, with the last term being \(376.5\), we rearranged the formula to find that the series consists of 150 terms.

• Knowing the number of terms is essential when calculating the sum of the entire series.
• It tells us the progression's extent, which can be essential for further mathematical or real-world applications.