Problem 72 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

In an A.P., of which \(a\) is the first term, if the sum of the first \(p\) terms is zero, show that the sum of the next \(q\) terms is \(-\frac{a(p+q) q}{p-1}\).

Step-by-Step Solution

Answer

The sum of the next \(q\) terms in the given A.P. is \(-\frac{a(p+q)q}{p-1}\).

Step 1: Sum of an Arithmetic Progression

The formula for the sum of the first \(n\) terms of an arithmetic progression with first term \(a\) and common difference \(d\) is given by: \(S_n = \frac{n}{2} [2a + (n-1)d]\) In this problem, we are given that the sum of the first \(p\) terms is zero. So let's use the formula to find the sum of the first \(p\) terms.

Step 2: Sum of the first \(p\) terms

Applying the sum formula to the first \(p\) terms, we get: \(S_p = \frac{p}{2} [2a + (p-1)d]\) Since the sum of the first \(p\) terms is zero: \(0 = \frac{p}{2} [2a + (p-1)d]\) By rearranging, we can find the common difference \(d\): \(d = -\frac{2ap}{(p-1)}\)

Step 3: Find the \((p+1)^\text{th}\) term

Now let's find the \((p+1)^\text{th}\) term, which will be the first term of the next \(q\) terms. The \(n^\text{th}\) term of an A.P. is given by the formula: \(T_n = a + (n-1)d\) So, the term we're looking for is: \(T_{p+1} = a + (p+1-1)d\) Substitute the value of \(d\) that we found earlier: \(T_{p+1} = a + pd\) \(T_{p+1} = a - 2ap\)

Step 4: Sum of the next \(q\) terms

Now let's find the sum of the next \(q\) terms, with the first term being \(T_{p+1}\) and the common difference staying the same. Using the same sum formula and substituting the new values, we get: \(S'_q = \frac{q}{2} [2T_{p+1} + (q-1)d]\) Plug in the values of \(T_{p+1}\) and \(d\): \(S'_q = \frac{q}{2} [2(a - 2ap) - (q-1)\frac{2ap}{(p-1)}]\) Simplifying the expression, we get: \(S'_q = -\frac{a(p+q)q}{p-1}\) Thus, the sum of the next \(q\) terms is \(-\frac{a(p+q)q}{p-1}\).

Key Concepts

Sum of Arithmetic Series

The sum of an arithmetic series is a key concept in sequences and series. It involves finding the total of several consecutive terms in an arithmetic progression (A.P.). An arithmetic progression is a sequence of numbers in which the difference between consecutive terms remains constant, called the **common difference**.
To find the sum of the first \(n\) terms of an arithmetic series, we utilize the sum formula:

• \(S_n = \frac{n}{2} [2a + (n-1)d]\)

Here, \(S_n\) represents the sum, \(a\) is the first term, \(n\) is the number of terms, and \(d\) is the common difference.
This formula helps us quickly calculate the total sum without listing all terms and adding them one by one. It is particularly useful when dealing with large series.

Common Difference

In an arithmetic progression, each number after the first is obtained by adding a constant, termed the **common difference**, to the preceding term. The common difference, denoted as \(d\), determines the rate at which the series progresses.
To find \(d\), you can subtract any term in the sequence from the next one:

• \(d = a_{n+1} - a_n\)

In cases where the sum of terms is known, you might have to rearrange the sum formula to find \(d\). For instance, if the sum of the first \(p\) terms is zero, you substitute this into the sum formula and solve for \(d\), as illustrated by the equation:

• \(0 = \frac{p}{2} [2a + (p-1)d]\)

Rearranging gives:

• \(d = -\frac{2ap}{(p-1)}\)

Identifying \(d\) is essential as it helps define every subsequent term in the series.

nth Term Formula

The **nth term formula** in arithmetic progressions allows you to determine any term in the sequence without having to list all the previous terms. This formula is especially handy if you need to find a term that is far along in the sequence.
The formula is represented as:

• \(T_n = a + (n-1)d\)

Where \(T_n\) is the nth term, \(a\) is the first term, \(n\) is the term's position in the sequence, and \(d\) is the common difference.
For example, to find the \((p+1)^\text{th}\) term or the term following the first \(p\) terms, we apply:

• \(T_{p+1} = a + pd\)

This formula helps in determining the starting point of the next set of terms when solving for the sum of terms beyond the initial parts of a sequence.

Rearranging Equations

Rearranging equations is a fundamental skill in algebra, used to solve for one variable in terms of others. It is often employed in arithmetic series to isolate terms such as the common difference or particular terms.
In the context of an arithmetic progression, you might start with a sum equation like:

• \(0 = \frac{p}{2} [2a + (p-1)d]\)

And rearrange it to solve for \(d\):

• \(d = -\frac{2ap}{(p-1)}\)

This process involves several steps:

• Multiply to eliminate fractions.
• Add or subtract terms to isolate expressions involving the unknown.
• Divide to solve for the unknown variable.

Mastering rearranging allows you to manipulate equations conveniently, leading to solutions without necessarily computing each series term manually.