Understanding the Arithmetic Progression (AP) Problem

The question involves an Arithmetic Progression (AP), which is a sequence of numbers such that the difference between the consecutive terms is constant. This constant difference is called the common difference, denoted by \(d\).

In an AP, the n-th term \(a_n\) can be expressed in terms of the first term \(a_1\) and the common difference \(d\) as:

\(a_n = a_1 + (n-1)d\)

We are given a specific sum involving several terms of an AP:

\(a_1 + a_5 + a_{10} + a_{15} + a_{20} + a_{25} + a_{30} + a_{34} = 300\)

We need to find the sum of the first 34 terms of this AP, which is \(\displaystyle \sum_{n=1}^{34} a_n\).

Utilizing AP Properties

A useful property of an AP of length \(N\) is that the sum of terms equidistant from the beginning and the end is constant. That is, for an AP \(a_1, a_2, \dots, a_N\), we have:

\(a_k + a_{N-k+1} = a_1 + a_N\)

In this problem, we are dealing with the first 34 terms, so \(N=34\). The property becomes:

\(a_k + a_{34-k+1} = a_k + a_{35-k} = a_1 + a_{34}\)

Let's examine the indices of the terms given in the sum:

\(a_1, a_5, a_{10}, a_{15}, a_{20}, a_{25}, a_{30}, a_{34}\)

We can pair these terms based on the property \(a_k + a_{35-k}\):

• \(a_1\) pairs with \(a_{34}\) because \(1+34 = 35\). So, \(a_1 + a_{34}\).
• \(a_5\) pairs with \(a_{30}\) because \(5+30 = 35\). So, \(a_5 + a_{30}\).
• \(a_{10}\) pairs with \(a_{25}\) because \(10+25 = 35\). So, \(a_{10} + a_{25}\).
• \(a_{15}\) pairs with \(a_{20}\) because \(15+20 = 35\). So, \(a_{15} + a_{20}\).

According to the AP property, each of these pairs sums to the same value, which is equal to the sum of the first and the last term:

\(a_1 + a_{34} = a_5 + a_{30} = a_{10} + a_{25} = a_{15} + a_{20}\)

Calculating the Sum of the First and Last Term

The given sum can be rewritten by grouping these pairs:

\((a_1 + a_{34}) + (a_5 + a_{30}) + (a_{10} + a_{25}) + (a_{15} + a_{20}) = 300\)

Since each parenthetical term is equal to \(a_1 + a_{34}\), we have:

\((a_1 + a_{34}) + (a_1 + a_{34}) + (a_1 + a_{34}) + (a_1 + a_{34}) = 300\)

\(4 \times (a_1 + a_{34}) = 300\)

Now, we can solve for the sum of the first and last term:

\(a_1 + a_{34} = \frac{300}{4}\)

\(a_1 + a_{34} = 75\)

Finding the Sum of the First 34 Terms

The sum of the first \(N\) terms of an AP, denoted by \(S_N\), is given by the formula:

\(S_N = \frac{N}{2}(a_1 + a_N)\)

In this case, we need to find the sum of the first 34 terms, so \(N=34\). Using the formula:

\(S_{34} = \frac{34}{2}(a_1 + a_{34})\)

We have already found that \(a_1 + a_{34} = 75\). Substitute this value into the formula:

\(S_{34} = \frac{34}{2}(75)\)

\(S_{34} = 17 \times 75\)

Let's calculate \(17 \times 75\):

So, the sum of the first 34 terms is 1275.

\(\displaystyle \sum_{n=1}^{34} a_n = S_{34} = 1275\)

Summary of Steps

• Identify the properties of an Arithmetic Progression, specifically the sum of terms equidistant from the start and end.
• Use the given sum \(a_1 + a_5 + a_{10} + a_{15} + a_{20} + a_{25} + a_{30} + a_{34} = 300\) and the property to find the value of \(a_1 + a_{34}\).
• Apply the formula for the sum of the first \(N\) terms of an AP, \(S_N = \frac{N}{2}(a_1 + a_N)\), with \(N=34\).
• Substitute the value of \(a_1 + a_{34}\) into the sum formula to find the required sum.

Revision Table: Key AP Concepts

Additional Information on Arithmetic Progressions

Arithmetic Progressions are fundamental sequences in mathematics. They appear in various applications, from simple counting patterns to more complex financial calculations. Understanding the properties of APs, such as the constant common difference and the relationships between terms, is crucial for solving problems related to sums and specific terms in the sequence.

The formula \(S_N = \frac{N}{2}(a_1 + a_N)\) is particularly useful when the first and last terms are known or can be easily found. The alternative formula \(S_N = \frac{N}{2}(2a_1 + (N-1)d)\) is used when the first term and common difference are known.

Problems involving sums of non-consecutive terms, as seen in this question, often leverage the property that the sum of terms equidistant from the beginning and end is constant. Identifying these pairs is key to simplifying the given information.