Question

Let \(b_{i}>1\) for \(i=1,2, \ldots, 101 .\) Suppose \(\log _{e} b_{1}, \log _{e} b_{2}, \ldots, \log _{e} b_{101}\) are in Arithmetic Progression (A.P.) with the common difference \(\log _{e} 2 .\) Suppose \(a_{1}, a_{2}, \ldots, a_{101}\) are in A.P. such that \(a_{1}=b_{1}\) and \(a_{51}=b_{51} .\) If \(t=b_{1}+b_{2}+\cdots+b_{51}\) and \(s=a_{1}+a_{2}+\cdots+a_{51}\), then (A) \(s>t\) and \(a_{101}>b_{101}\) (B) \(s>t\) and \(a_{101}b_{101}\) (D) \(s

Short answer

Since the sum of the terms of \(b_i\) (\textbf{t}) grows exponentially and the sum of the terms of \(a_i\) (\textbf{s}) grows linearly, it follows that \(s<t\). Moreover, \(a_{101} < b_{101}\) because the exponential increase of terms in \(b_i\) will always surpass the linear increase of terms in \(a_i\). Therefore, the correct answer is (D) \(s<t\) and \(a_{101}<b_{101}\).

Step-by-step solution

Understanding the Arithmetic Progression of Logarithms

Given that \(\log_e b_i\) are in an arithmetic progression, this means that each term after the first is formed by adding \(\log_e 2\) to the previous term. We can express subsequent terms of the sequence as \(\log_e b_1 + (i-1)\log_e 2\) for \(i = 1, 2, ..., 101\).

Representing the Terms of Sequence \(a_i\)

The terms \(a_1, a_2, \.\.\.\), \(a_{101}\) are also in an arithmetic progression with the same first term and 51st term as sequence \(b_i\). Thus their common difference can be found using the formula for the \(n\)th term of an A.P.: \(a_n = a_1 + (n-1)d\), where \(a_{51} = b_{51} = a_1 + 50d\). We need to find the common difference \(d\) for the sequence \(a_i\).

Calculating the Common Difference for \(a_i\)

Since \(a_1 = b_1\) and \(a_{51} = b_{51}\), and we know \(b_{51} = b_1 \cdot (e^{\log_e 2})^{50}\), we can set up the equation \(b_1 \cdot (e^{50 \log_e 2}) = a_1 + 50d\) which simplifies to \(b_1 \cdot 2^{50} = b_1 + 50d\). By dividing both sides by \(b_1\), we get \(2^{50} = 1 + \frac{50d}{b_1}\), which allows us to solve for \(d\), the common difference of \(a_i\).

Comparing Sums \(t\) and \(s\)

The sum of the first 51 terms of the series \(b_i\), denoted \(t\), is greater than the sum \(s\) of the first 51 terms of series \(a_i\) since the terms of \(b_i\) grow exponentially due to multiplication by \(2\), while the terms of \(a_i\) grow linearly. Accordingly, we can infer that \(s < t\).

Determining the Relationship between \(a_{101}\) and \(b_{101}\)

The 101st term of \(b_i\), \(b_{101}\), is found by the same process as earlier, \(b_1 \cdot (e^{100 \log_e 2}) = b_1 \cdot 2^{100}\), while \(a_{101}\) is \(a_1 + 100d\). Since exponential growth is faster than linear growth, \(b_{101}\) will be greater than \(a_{101}\). Thus, \(a_{101} < b_{101}\).

Key concepts

Logarithm Properties

Understanding logarithm properties is essential for solving many mathematical problems, including those dealing with arithmetic progressions where the terms are logarithms. A logarithm, specifically the natural logarithm denoted as \( \log_e \), represents the power to which a number, known as the base (in this case \(e\), the Euler's number), must be raised to obtain another number. One of the fundamental properties of logarithms is that the logarithm of a product is the sum of the logarithms. Mathematically, this is expressed as \( \log_e(x \times y) = \log_e x + \log_e y \).

Another crucial property is that the logarithm of a power is the exponent times the logarithm, which is \( \log_e(x^n) = n \log_e x \). This property is particularly relevant to the exercise, as it shows how we can convert an exponential sequence into an arithmetic sequence via logarithms. When we apply this property, it helps us understand why a sequence \( \log_e b_i \) with a common difference \( \log_e 2 \) relates to an underlying sequence where each term is twice the previous term. These properties of logarithms are tools that empower students to transform and solve complex equations more easily.

Exponential Growth

Exponential growth occurs when the rate of increase of a quantity is proportional to its current value. In the context of the given arithmetic progression exercise, it is implied when the terms \( b_i \) are effectively multiplied by a constant factor with each step. Here, it's critical to realize that exponential growth is distinctly different from linear growth, which is the type of growth that occurs in an arithmetic sequence.

In real-world scenarios, exponential growth can be seen in things like compound interest, population growth, and certain types of viral spread. It's characteristically rapid and accelerating, which was reflected in the solution where \( b_{101} \) is exponentially larger than \( a_{101} \) because it grows by repeatedly doubling (due to the common difference \( \log_e 2 \) in the logarithmic expression) rather than by incremental, fixed additions.

Common Difference in Arithmetic Sequence

The concept of the common difference in an arithmetic sequence is integral to identifying and describing the sequence's pattern. The common difference \(d\) is the constant amount that each term in the sequence changes by when moving from one term to the next. For instance, if we have an arithmetic sequence where each term increases by 3, we would express this as \(d = 3\).

In the step by step solution, finding the common difference for the sequence \( a_i \) played a crucial role, as it brought us closer to comparing the sums \( s \) and \( t \) and eventually determining the relationship between \( a_{101} \) and \( b_{101}\). When students grasp the concept of the common difference, they can apply it to find not just any term in the sequence but also the sum of a given number of terms. This concept illustrates a fundamental difference between linear growth (seen in arithmetic sequences) and exponential growth (demonstrated by the sequence \( b_i \)), a distinction that proved crucial to solving the problem and will be encountered frequently in various mathematical contexts.