Problem 64 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

Show that the sum of an A.P. whose first term is \(a\), second term is \(b\) and the last term is \(c\) is equal to \(\frac{(a+c)(b+c-2 a)}{2(b-a)}\)

Step-by-Step Solution

Answer

The sum of an A.P. whose first term is \(a\), second term is \(b\), and last term is \(c\) can be found using the formula \(S_n = \frac{n}{2}(a_1 + a_n)\). By identifying the common difference (\(d = b - a\)) and the total number of terms (\(n = \frac{(b+c-2a)}{b-a} + 1\)), we can apply the A.P. sum formula to obtain \(S_n = \frac{\left(\frac{(b+c-2a)}{b-a} + 1\right)}{2}(a + c)\). Simplifying the expression, we end up with the given formula, \(\frac{(a+c)(b+c-2a)}{2(b-a)}\), thus proving it.

Step 1: Identify the Common Difference

The first term of the A.P. is given as \(a\), and the second term is given as \(b\). The common difference is the difference between these two terms: \(d = b - a\).

Step 2: Identify the Total Number of Terms

The last term of the A.P. is given as \(c\). We can express the last term in terms of the first term and the common difference as \(c = a + (n-1)d\), where \(n\) is the total number of terms in the series. We plug in the common difference \(d = b - a\) and rearrange to solve for \(n\): \(c = a + (n-1)(b-a)\) Since we want to find the total number of terms \(n\), we can rearrange the equation as follows: \(n = \frac{c - a + b - a}{b - a} + 1 = \frac{(b+c-2a)}{b-a} + 1\)

Step 3: Apply the Sum of an A.P. Formula

The sum of an A.P. series is given by the formula \(S_n = \frac{n}{2}(a_1 + a_n)\), where \(S_n\) is the sum of the series with \(n\) terms, \(a_1\) is the first term, and \(a_n\) is the last term. Now, we will plug in the values we found in steps 1 and 2 to find the sum of the A.P. series: \(S_n = \frac{\left(\frac{(b+c-2a)}{b-a} + 1\right)}{2}(a + c)\)

Step 4: Simplify and Prove the Given Formula

We will now simplify the expression to prove the given formula: \(S_n = \frac{(b+c-2a) + (b-a)}{2(b-a)}\cdot (a+c)\) \(LHS = \frac{(b+c-2a)+(b-a)}{2(b-a)} \cdot (a+c)\) \(=\frac{(a+c)(b+c-2a)}{2(b-a)}\) As we can see, the sum of the A.P. simplifies to the given formula, \(\frac{(a+c)(b+c-2a)}{2(b-a)}\). Thus, we have proven the given formula.

Key Concepts

Common Difference

In an arithmetic progression (A.P.), the pattern of the series is determined by the common difference. This difference, denoted as \(d\), is the amount by which each term increases or decreases from the previous one.

• To find the common difference, simply subtract the first term from the second term.
• For example, if the first term \( a \) is 3 and the second term \( b \) is 7, then \( d = b - a = 7 - 3 = 4 \).
• This consistent difference allows us to determine any term in the sequence if we know the starting value and the number of steps along the progression.

Understanding this foundational concept is crucial because the common difference helps in identifying the nature of the sequence and calculating other elements like the sum or total number of terms.

Sum of an Arithmetic Series

Finding the sum of an arithmetic series involves adding all the terms together. The formula to calculate this sum, \( S_n \), is: \[ S_n = \frac{n}{2} (a_1 + a_n) \] where \( n \) is the number of terms, \( a_1 \) is the first term, and \( a_n \) is the last term.

• This formula applies because an A.P. forms a symmetrical pattern that can be paired to simplify calculations. Each opposite pair (first and last, second and second-last, etc.) adds up to the same total.
• Let's consider an example where the first term \( a_1 \) is 2, and the last term \( a_n \) is 10 in a series with 5 terms. Plug these into the formula: \( S_5 = \frac{5}{2} (2 + 10) = \frac{5}{2} \times 12 = 30 \).

The given sum formula derived in the exercise is specific to the sequence parameters presented, emphasizing how these terms are algebraically manipulated to arrive at a description like \( \frac{(a+c)(b+c-2a)}{2(b-a)} \). By using the sum formula, we calculate how the addition of each term constructed using the common difference accumulates.

Number of Terms in an A.P.

Determining the number of terms in an arithmetic progression is vital for understanding its scope. We use the last term formula: \[ c = a + (n-1)d \] where \( c \) is the last term, \( a \) is the first term, \( d \) is the common difference, and \( n \) is the number of terms. Rearranging this formula gives us the number of terms: \[ n = \frac{c-a}{d} + 1 \]

• This equation can help identify \( n \) when the initial, ending values, and the common difference are known.
• For example, if \( a = 2 \), \( c = 14 \), and \( d = 3 \), then \( n = \frac{14 - 2}{3} + 1 = 5 \).
• In the problem's context, substituting the common difference \( d = b-a \) allows solving for \( n \) as shown in the step-by-step solution.

This count of terms is particularly useful when applying the sum formula, ensuring all elements have been accounted for efficiently. It underscores how arithmetic progressions are predictable, making such calculations systematic and straightforward.