Problem 22 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

Find the number of terms in the series \(20+19 \frac{1}{3}+18 \frac{2}{3}+\cdots \cdots\) of which the sum is 300 . Explain the double answer. Also find the maximum sum of the series.

Step-by-Step Solution

Answer

The number of terms in the series is 60. The double answer originates from two potential values of n from the quadratic equation, but only one of them, 60, makes sense in the context. The maximum sum of the series is -580.

Step 1: Identify the pattern in the series

Observe the given series: \(20, 19\frac{1}{3}, 18\frac{2}{3}, \cdots\) The first term is an integer, and all subsequent terms are mixed fractions (a combination of an integer and a proper fraction). The difference between two consecutive terms is \(\frac{-1}{3}\). We can now use this to form a general expression for the series.

Step 2: Find the general expression

Let the nth term of the series (an) be expressed as: \(a_n = a_{1} + (n-1)d\) where, \(a_{1}\) is the first term, which is \(20\), n is the number of terms, d is the common difference between terms, which is \(\frac{-1}{3}\). So, the nth term can be expressed as: \(a_{n} = 20 + (n-1)\frac{-1}{3}\)

Step 3: Use the sum formula to find the number of terms

The sum of an arithmetic series, denoted by S, can be calculated using the formula: \(S_{n} = \frac{n}{2}[2a_{1} + (n-1)d]\) Here, the sum is given as 300. Thus, \(300 = \frac{n}{2}[2(20) + (n-1)\frac{-1}{3}]\) Now we need to solve for n.

Step 4: Solve for n and Explain the double answer

Expanding the equation from Step 3, we get: \(600 = n[40 - (n-1)\frac{1}{3}]\) In order to solve for n, let's set: \(x = (n - 1) \Longrightarrow n = x + 1\) So the equation becomes: \(600 = (x + 1)(40 - \frac{x}{3})\) This is a quadratic equation in x. Expand and solve for x: \(600 - 40x - \frac{x^{2}}{3} + \frac{x}{3} = 0\) Multiplying the entire equation by 3 to eliminate fractions: \(1800 - 120x - x^{2} + x = 0\) Rearrange and simplify: \(x^{2} - 121x + 1800 = 0\) Now, apply the quadratic formula, remembering that n = x + 1: \(n = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}\) where a = 1, b = -121, and c = 1800. We get two possible values for n: \(n_{1} = 60\) \(n_{2} = -30\) Since the number of terms cannot be negative, we ignore n2, and thus, we have n1 = 60 terms in the series. The double answer comes from the quadratic equation leading to two potential values of n, i.e., 60 and -30, but only one of them, 60, has a meaningful interpretation in this context.

Step 5: Find the maximum sum of the series

To find the maximum sum of the series, we use the sum formula again: \(S_{n} = \frac{n}{2}[2a_{1} + (n-1)d]\) Using the number of terms (n = 60) and substituting the values: \(S_{60} = \frac{60}{2}[2(20) + (60-1)\frac{-1}{3}]\) Calculate the maximum sum: \(S_{60} = 30[40 - 59\frac{1}{3}]\) \(S_{60} = 30\left[-19\frac{1}{3}\right]\) The maximum sum of the series is -580.

Key Concepts

Common Difference

In an arithmetic series, the **common difference** between consecutive terms is the number that is either added or subtracted to create the sequence. It's a critical factor because it determines whether each term in the sequence will increase or decrease. In the exercise presented, the common difference is negative, specifically \(-\frac{1}{3}\). This means each term is \(-\frac{1}{3}\) smaller than the preceding term.

To establish the common difference, examine any two successive terms in the series. Subtract the latter from the former. If this operation returns the same value for each pair, that consistent result is your common difference.

For example, in the series given:

• First term: 20
• Second term: 19rac{1}{3}

Subtract the second term from the first: \(20 - 19 \frac{1}{3} = \frac{2}{3} \), but considering the subtraction as a downward operation for the sequence, it's \( - \frac{1}{3} \). This reflects the consistent pattern throughout the sequence.

Sum of Series

The sum of an arithmetic series is the total of all its terms. This sum can be efficiently calculated using a specific formula rather than adding each term directly. For the series provided, the sum formula is:

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

Here, \(n\) is the total number of terms you want to sum, \(a_1\) is the first term of the series, and \(d\) is the common difference.

In this exercise, with a series sum set as 300, you use the sum formula to find \(n\): the number of terms needed to reach that sum.

Once solved, such formulas can also determine maximum values and trends over specific intervals, like calculating the series' maximum sum when given its length (such as 60 terms, in this problem).

While direct addition might seem simple, leveraging the sum formula saves time and ensures accuracy, especially as sequences grow.

Quadratic Equation

A **quadratic equation** is any equation that takes the standard form \(ax^2 + bx + c = 0\). These equations can describe various physical phenomena and are solved using methods such as factoring, completing the square, or the quadratic formula.

The quadratic formula is:
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

In the given exercise, after setting up an expression to determine the number of terms ( ext{n}), a quadratic equation was derived for calculating the possible values. This arrived when expanding the equation embedded in the arithmetic sum.

Solving this quadratic presented two potential solutions, \(n_1\) and \(n_2\), derived from modifying the sequence's length through its arithmetic nature. Nevertheless, only one result -60 -made sense contextually, revealing how quadratic solutions can sometimes involve discarded extraneous results.

Understanding quadratic equations is vital, as they can arise in evaluating series and give insight into possible solutions.