Finding the Sum of an Arithmetic Progression (AP)

This problem asks us to find the sum of the first 50 terms of an Arithmetic Progression (AP). We are given information about two specific terms in the sequence: the 5th term and the 10th term. To find the sum of the first 50 terms, we first need to determine the first term (let's call it 'a') and the common difference (let's call it 'd') of the AP.

Understanding Arithmetic Progressions (AP)

An Arithmetic Progression is a sequence of numbers such that the difference between the consecutive terms is constant. This constant difference is known as the common difference, 'd'.

The formula for the \(n\)th term of an AP is:

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

Where:

• \(a_n\) is the \(n\)th term
• \(a\) is the first term
• \(n\) is the term number
• \(d\) is the common difference

The formula for the sum of the first \(n\) terms of an AP is:

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

or

\(S_n = \frac{n}{2}[a + a_n]\) (if the last term \(a_n\) is known)

Determining the First Term (a) and Common Difference (d)

We are given the 5th term and the 10th term:

• 5th term (\(n=5\)): \(a_5 = \frac{1}{10}\)
• 10th term (\(n=10\)): \(a_{10} = \frac{1}{5}\)

Using the formula \(a_n = a + (n-1)d\), we can write two equations:

1. For the 5th term: \(a_5 = a + (5-1)d \Rightarrow a + 4d = \frac{1}{10}\) (Equation 1)
2. For the 10th term: \(a_{10} = a + (10-1)d \Rightarrow a + 9d = \frac{1}{5}\) (Equation 2)

Now we have a system of two linear equations with two variables, 'a' and 'd'. We can solve this system to find the values of 'a' and 'd'. A simple way is to subtract Equation 1 from Equation 2:

\((a + 9d) - (a + 4d) = \frac{1}{5} - \frac{1}{10}\)

\(a + 9d - a - 4d = \frac{2}{10} - \frac{1}{10}\)

\(5d = \frac{1}{10}\)

Divide by 5 to find 'd':

\(d = \frac{1}{10 \times 5} = \frac{1}{50}\)

Now that we have the value of 'd', we can substitute it back into either Equation 1 or Equation 2 to find 'a'. Let's use Equation 1:

\(a + 4d = \frac{1}{10}\)

\(a + 4 \left(\frac{1}{50}\right) = \frac{1}{10}\)

\(a + \frac{4}{50} = \frac{1}{10}\)

\(a = \frac{1}{10} - \frac{4}{50}\)

To subtract these fractions, find a common denominator, which is 50:

\(a = \frac{5}{50} - \frac{4}{50}\)

\(a = \frac{5-4}{50} = \frac{1}{50}\)

So, the first term \(a = \frac{1}{50}\) and the common difference \(d = \frac{1}{50}\).

Calculating the Sum of the First 50 Terms

We need to find the sum of the first 50 terms, which is \(S_{50}\). We use the formula \(S_n = \frac{n}{2}[2a + (n-1)d]\) with \(n=50\), \(a=\frac{1}{50}\), and \(d=\frac{1}{50}\).

\(S_{50} = \frac{50}{2}\left[2\left(\frac{1}{50}\right) + (50-1)\left(\frac{1}{50}\right)\right]\)

\(S_{50} = 25\left[\frac{2}{50} + 49\left(\frac{1}{50}\right)\right]\)

\(S_{50} = 25\left[\frac{2}{50} + \frac{49}{50}\right]\)

\(S_{50} = 25\left[\frac{2+49}{50}\right]\)

\(S_{50} = 25\left[\frac{51}{50}\right]\)

Now, perform the multiplication:

\(S_{50} = \frac{25 \times 51}{50}\)

We can simplify this by cancelling out 25 from the numerator and the denominator:

\(S_{50} = \frac{51}{2}\)

Converting the fraction to a decimal:

\(S_{50} = 25.5\)

The sum of the first 50 terms is 25.5.

Comparing this result with the given options, the value 25.5 is represented by option 2, which is 25⋅5.

Summary of Calculations

Revision Table: Key Concepts in AP

Additional Information: Solving AP Problems

When solving problems involving Arithmetic Progressions, it's often helpful to follow these steps:

• Identify what is given (e.g., specific terms, number of terms, sum).
• Identify what needs to be found (e.g., a, d, \(a_n\), \(S_n\), n).
• Use the formulas for the \(n\)th term and the sum of \(n\) terms to set up equations based on the given information.
• Solve the equations (often a system of linear equations) to find the unknown variables, typically 'a' and 'd'.
• Use the calculated 'a' and 'd' values to find the required term or sum.
• Always double-check calculations, especially with fractions.

Understanding the relationship between the term number, the first term, the common difference, and the term's value is crucial for solving AP problems effectively.