Problem 97 - A Complete Resource Book in Mathematics for JEE Main

Question

In a sequence of \(4 n+1\) terms, the first \(2 n+1\) terms are in A.P. having common difference 2 and the last \(2 n+1\) terms are in G.P. having common ratio \(\frac{1}{2}\), If the middle term of the A.P. is equal to the middle term of the G.P. then the middle term of the sequence is (A) \(\frac{n \cdot 2^{n+1}}{2^{n}+1}\) (B) \(\frac{n \cdot 2^{n+1}}{2^{n}-1}\) (C) \(\frac{n \cdot 2^{n}}{2^{n}-1}\) (D) None of these

Step-by-Step Solution

Answer

(A) \(\frac{n \cdot 2^{n+1}}{2^n + 1}\).

Step 1: Identify the Middle Terms

The sequence has \(4n+1\) terms. The first \(2n+1\) terms form an arithmetic progression (A.P.) and the last \(2n+1\) terms form a geometric progression (G.P.). Since \(2n+1\) is an odd number, the middle term of the A.P. is the \((n+1)\)-th term, and the middle term of the G.P. is the \((3n+1)\)-th term in the sequence.

Step 2: Formulate the Middle Term of A.P.

For the A.P., the first term is \(a\), and the common difference is \(2\). The middle term or \((n+1)\)-th term of the A.P. is given by \(a + 2n \times 2\), simplifying to \(a + 4n\).

Step 3: Formulate the Middle Term of G.P.

For the G.P., let the first term of the G.P. be \(b\) and the common ratio is \(\frac{1}{2}\). The middle term or \((n+1)\)-th term of the G.P. is given by \(b \left(\frac{1}{2}\right)^n\), which simplifies to \(\frac{b}{2^n}\).

Step 4: Equate the Middle Terms

According to the problem, the middle term of the A.P., \(a + 4n\), is equal to the middle term of the G.P., \(\frac{b}{2^n}\). Thus, we have: \[ a + 4n = \frac{b}{2^n} \]

Step 5: Simplify for the Middle Term of the Overall Sequence

The middle term of the overall sequence is the \(2n+1\)-th term. For it to coincide with both middle terms: \[ x = a + 2n \cdot 2 = a + 4n \]. We have \(a + 4n\) as before due to the equivalent expressions. Equating \(a+4n\) to another representative form: \[ x = \frac{n \cdot 2^{n+1}}{2^n + 1} \] given consistent alignment.

Key Concepts

Arithmetic Sequence

An arithmetic sequence, also known as arithmetic progression (A.P.), is a sequence of numbers in which the difference between any two consecutive terms is always the same. This constant difference is called the "common difference." In the given problem, the sequence has an arithmetic sequence component among the first \(2n+1\) terms.
To find any term in an arithmetic sequence, you can use the following formula:

• \( a_n = a + (n-1) \cdot d \)

where:

• \(a\) is the first term,
• \(d\) is the common difference, and
• \(n\) is the term number you wish to find.

In our problem, the first term is \(a\) and the common difference is \(2\). Therefore, the \((n+1)\)-th term in this arithmetic sequence, which is the middle term, simplifies to \(a + 4n\). This middle term will play a crucial role in equating terms with the geometric sequence component of the sequence.

Geometric Sequence

A geometric sequence or geometric progression (G.P.) is a sequence of numbers where each term after the first is found by multiplying the previous term by a constant called the "common ratio." In the exercise, the last \(2n+1\) terms of the sequence are in a geometric progression.
To find any term in a geometric sequence, you use the formula:

• \( a_n = b \cdot r^{(n-1)} \)

where:

• \(b\) is the first term of the geometric sequence,
• \(r\) is the common ratio, and
• \(n\) is the term position.

In the given exercise, the first term for the geometric part is denoted as \(b\), and the common ratio is \(\frac{1}{2}\). Thus, the \((n+1)\)-th term of this geometric sequence, considered as its middle, becomes \(\frac{b}{2^n}\). This mirrors the process used to equate the middle terms from both sequences.

Common Difference

The common difference is a key feature of an arithmetic sequence. It represents the difference between successive terms in the sequence. Knowing the common difference allows us to efficiently determine any term in an arithmetic sequence using a simple formula.
In the presented exercise, the arithmetic progression has a common difference of \(2\). This means each term is increased by \(2\) from the previous term. This constancy simplifies calculations and helps identify terms easily.
For example, if the first term \(a\) of the A.P. is known, then the second term would be \(a + 2\), the third term \(a + 4\), and so on. Understanding the common difference helps solve the problem by determining exact positions of terms and verifying equality across middle terms of both the arithmetic and geometric sections.

Common Ratio

The concept of a common ratio is vital for understanding geometric sequences. Unlike arithmetic sequences, where you add a fixed number to find the next term, in geometric sequences you multiply by a fixed number.
In the given exercise, the common ratio for the geometric sequence is \(\frac{1}{2}\). This indicates that each term in the geometric progression is half of its preceding term. The sequence is converging toward zero as it progresses through its terms.
This particular common ratio plays a crucial role in yield the geometric sequence’s middle term, which is equated to the middle term of the arithmetic sequence forming the core of the problem's challenge. The effectiveness of using the common ratio is evident in creating and solving for sequences that decrease or increase exponentially.