Problem 76 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

The sums of \(n, 2 n, 3 n\) terms of an A.P. are \(S_{1}, S_{2}, S_{3}\) respectively. Prove that \(S_{3}=3\left(S_{2}-S_{1}\right)\).

Step-by-Step Solution

Answer

The proof that \(S_3=3\left(S_2-S_1\right)\) for the sums of n, 2n, and 3n terms of an A.P. can be obtained by first finding the general formula for the sum of n terms of an A.P., which is: \( S_n = n \left(\frac{2a + (n-1)d}{2}\right)\). Then, we express the given sums \(S_1, S_2, S_3\) in terms of a and d, and simplify the expressions. Finally, we prove the relationship \(S_3=3\left(S_2-S_1\right)\) by calculating and simplifying \(3\left(S_2-S_1\right)\) and showing that it is equal to the expression we obtained for \(S_3\). Therefore, the proof is complete: \[ S_{3}=3\left(S_{2}-S_{1}\right) \]

Step 1: Sum of n terms of an A.P. formula

The sum of n terms of an A.P. with first term a and common difference d can be calculated using the following formula: \[ S_n = n \left(\frac{2a + (n-1)d}{2}\right)\]

Step 2: Express the given sums in terms of a and d

We are given the sums \(S_1, S_2, S_3\) for n, 2n, 3n terms, respectively. Using the formula from step 1, let's write these sums: \[ S_1 = n \left(\frac{2a + (n-1)d}{2}\right)\] \[ S_2 = 2n \left(\frac{2a + (2n-1)d}{2}\right)\] \[ S_3 = 3n \left(\frac{2a + (3n-1)d}{2}\right)\]

Step 3: Simplify the expressions

Now, we will simplify these expressions to make them more manageable: \[ S_1 = n\left(a + \frac{(n-1)d}{2}\right)\] \[ S_2 = n\left(a + (n-\frac{1}{2})d\right)\] \[ S_3 = n\left(a + (n-\frac{1}{3})d\right)\]

Step 4: Prove the relationship

We have to prove that \(S_3=3\left(S_2-S_1\right)\). Let's use the expressions obtained in step 3 to calculate \(3\left(S_2-S_1\right)\): \[ 3\left(S_2-S_1\right) = 3n\left(\left(a + (n-\frac{1}{2})d\right) - \left(a + \frac{(n-1)d}{2}\right)\right) \]

Step 5: Simplify the expression

Simplify the expression obtained in step 4: \begin{align*} 3\left(S_2-S_1\right) &= 3n\left((n - \frac{d}{2}) - \frac{(n-1)d}{2}\right) \\ &= 3n\left(n - \frac{1}{3}d\right) \\ \end{align*} Notice that the expression we obtained for \(3\left(S_2-S_1\right)\) is equal to what we obtained for \(S_3\) in step 3. Therefore, it's proven that: \[ S_{3}=3\left(S_{2}-S_{1}\right) \]

Key Concepts

sum of n terms

In an arithmetic progression (A.P.), we often need to calculate the sum of the first \(n\) terms. Understanding how to find this sum is crucial. The formula to find the sum \(S_n\) of the first \(n\) terms of an A.P. is given by:\[ S_n = n \left(\frac{2a + (n-1)d}{2}\right) \]Here, \(a\) is the first term, \(d\) is the common difference, and \(n\) is the number of terms. This formula arises because the arithmetic progression is symmetrical, so averaging the first and the last term, and multiplying by the number of terms, gives us the total sum.
This concept helps in solving problems where you need to find how much a series accumulates over a given number of terms. It is also used extensively in various real-life scenarios, like calculating savings or installments over time.

common difference

The common difference \(d\) in an arithmetic progression is a pivotal part of understanding the sequence. It is the constant amount that each term in the series increases or decreases by compared to the previous term.
To find the common difference, subtract the first term of the A.P. from the second term, or any term from the preceding term:

• If the sequence is increasing, \(d\) is positive.
• If the sequence is decreasing, \(d\) is negative.

Knowing the common difference allows you to identify the pattern in the series. This is important for writing the formulae used to find specific terms or the sum of terms in the sequence, as reflected in the formulas used in our problem.

first term

The first term \(a\) of an arithmetic progression is the initial term in the sequence, starting point. In our formula for the sum of \(n\) terms \(S_n\), and when calculating specific terms, \(a\) is crucial.

• It forms the foundation upon which future terms are built.
• Every other term in the A.P. is derived by adding the common difference to the first term, iteratively.

Understanding the first term is essential because it dictates the sequence's initial position and affects every subsequent calculation, like the sums or finding the nth term. It is valuable for anyone analyzing how a series will progress over time.

series formula

The series formula for an arithmetic progression allows us to succinctly describe the progression’s property over \(n\) terms. When you're calculating sums or individual terms, the series formula compacts this into meaningful expressions that can be solved methodically.The general nth-term formula is:\[ a_n = a + (n-1)d \]This formula lets you compute any term in the sequence using the first term \(a\) and the common difference \(d\). For the sum of \(n\) terms, the series formula expands on this:\[ S_n = n \left(\frac{2a + (n-1)d}{2}\right) \]These formulas are potent tools for solving A.P. problems efficiently. They encapsulate the growth of the sequence for easy computation of any term or total sum over specific intervals.