Understanding the Arithmetic Progression Problem

The question asks us to find the sum of the first ten terms of an arithmetic progression (AP). We are given the first term and a condition relating the sum of the first five terms to the sum of the next five terms.

Given Information

• First term of the AP ($a_1$) = 2
• The sum of the first five terms ($S_5$) is equal to one-fourth of the sum of the next five terms. The sum of the next five terms (terms 6 through 10) can be represented as the total sum of the first ten terms ($S_{10}$) minus the sum of the first five terms ($S_5$).

Setting up the Relationship

Let $S_n$ denote the sum of the first $n$ terms of an AP. The sum of the next five terms (from the 6th to the 10th term) is $S_{10} - S_5$.

According to the problem statement, we have the relationship:

\( S_5 = \frac{1}{4}(S_{10} - S_5) \)

We can rearrange this equation to simplify it:

\( 4S_5 = S_{10} - S_5 \)

\( 4S_5 + S_5 = S_{10} \)

\( 5S_5 = S_{10} \)

Formulas for Sum of an AP

The formula for the sum of the first $n$ terms of an arithmetic progression is:

\( S_n = \frac{n}{2}[2a_1 + (n-1)d] \)

where \(a_1\) is the first term and \(d\) is the common difference.

Using this formula, we can write the expressions for \(S_5\) and \(S_{10}\):

• For \(n=5\): \( S_5 = \frac{5}{2}[2a_1 + (5-1)d] = \frac{5}{2}[2a_1 + 4d] \)
• For \(n=10\): \( S_{10} = \frac{10}{2}[2a_1 + (10-1)d] = 5[2a_1 + 9d] \)

Finding the Common Difference (d)

We know \(a_1 = 2\) and \(5S_5 = S_{10}\). Let's substitute the expressions for \(S_5\) and \(S_{10}\) into this relationship:

\( 5 \times \frac{5}{2}[2a_1 + 4d] = 5[2a_1 + 9d] \)

Substitute \(a_1 = 2\):

\( 5 \times \frac{5}{2}[2(2) + 4d] = 5[2(2) + 9d] \)

\( \frac{25}{2}[4 + 4d] = 5[4 + 9d] \)

We can divide both sides by 5:

\( \frac{5}{2}[4 + 4d] = [4 + 9d] \)

Multiply both sides by 2:

\( 5[4 + 4d] = 2[4 + 9d] \)

Distribute the numbers:

\( 20 + 20d = 8 + 18d \)

Now, solve for \(d\). Subtract \(18d\) from both sides:

\( 20 + 20d - 18d = 8 \)

\( 20 + 2d = 8 \)

Subtract 20 from both sides:

\( 2d = 8 - 20 \)

\( 2d = -12 \)

Divide by 2:

\( d = \frac{-12}{2} \)

\( d = -6 \)

The common difference of the AP is -6.

Calculating the Sum of the First Ten Terms (\(S_{10}\))

Now that we have the first term \(a_1 = 2\) and the common difference \(d = -6\), we can find the sum of the first ten terms using the formula \( S_{10} = 5[2a_1 + 9d] \):

\( S_{10} = 5[2(2) + 9(-6)] \)

\( S_{10} = 5[4 - 54] \)

\( S_{10} = 5[-50] \)

\( S_{10} = -250 \)

The sum of the first ten terms of the AP is -250.

Revision Table: Arithmetic Progression Formulas

Additional Information on Arithmetic Progressions

An arithmetic progression (AP) 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 \(d\).

• Identifying an AP: Check if the difference between any two consecutive terms is the same throughout the sequence. For example, 2, -4, -10, -16, ... is an AP with \(a_1=2\) and \(d=-6\).
• Common Difference: \(d = a_n - a_{n-1}\) for any \(n > 1\).
• Properties of APs:
  • If three terms are in AP, say \(a, b, c\), then \(2b = a + c\). This means the middle term is the arithmetic mean of the other two.
  • If a constant is added to or subtracted from each term of an AP, the resulting sequence is also an AP with the same common difference.
  • If each term of an AP is multiplied or divided by a non-zero constant, the resulting sequence is also an AP. The new common difference is the original common difference multiplied or divided by the constant.
• Arithmetic Mean: For any two numbers \(a\) and \(b\), their arithmetic mean is \(\frac{a+b}{2}\). If a number is the arithmetic mean of two other numbers, the three numbers form an AP.

Understanding these concepts helps in solving various problems related to arithmetic progressions, including finding terms, sums, or missing values within the sequence.