Let bi 1 for i = 1, 2, ..., 101. Suppose loge b1, loge b2, ..., loge b101 are in Arithmetic Progre

Q: Wait, why do we assume $b_i$ is a Geometric Progression just because its log is an A.P.?

That is a fundamental property of logarithms! If $\log_e b_i$ forms an A.P., then $\log_e b_{i+1} - \log_e b_i = d$. By the laws of logs, this is $\log_e(b_{i+1}/b_i) = d$, which means $b_{i+1}/b_i = e^d$. Since $e^d$ is a constant, the ratio between consecutive terms is constant, which is the definition of a G.P.

Q: Why can't we just use energy conservation or some physical analogy here?

While physics problems often use energy conservation, this is a pure mathematical sequence problem. However, you can think of it as 'growth rates.' The G.P. has a constant percentage growth, while the A.P. has a constant absolute growth. The exponential growth will eventually outpace the linear growth, which is the core intuition behind the comparison.

Q: How do I know for sure that the A.P. sum is greater than the G.P. sum between $i=1$ and $i=51$?

Think of the graph. A convex function (like $2^x$) always lies below the secant line connecting two points on its curve. Since $a_i$ is the line connecting $(1, b_1)$ and $(51, b_{51})$, and $b_i$ is the convex curve, the line must stay above the curve for all $i$ between 1 and 51. Thus, the sum of the line's values must be greater.

Q: Is it always true that $a_{101} < b_{101}$?

Yes. Once you move past the second intersection point ($i=51$), the exponential function $b_i$ grows much faster than the linear function $a_i$. By the time you reach $i=101$, the exponential term $2^{100}$ completely dwarfs the linear term $2^{51}-1$.

Q: Do I need to calculate the exact value of $b_1$ to solve this?

Not at all! Notice that $b_1$ appears as a common factor in all our expressions for $s, t, a_{101},$ and $b_{101}$. Since we are given $b_1 > 1$, it acts as a scaling factor that doesn't change the direction of the inequalities. You can treat it as a positive constant.

Question

Let $b_i > 1$ for $i = 1, 2, \dots, 101$. Suppose $\log_e b_1, \log_e b_2, \dots, \log_e b_{101}$ are in Arithmetic Progression (A.P.) with the common difference $\log_e 2$. Suppose $a_1, a_2, \dots, a_{101}$ are in A.P. such that $a_1 = b_1$ and $a_{51} = b_{51}$. If $t = b_1 + b_2 + \dots + b_{51}$ and $s = a_1 + a_2 + \dots + a_{51}$, then Options: (A) $s > t$ and $a_{101} > b_{101}$ (B) $s > t$ and $a_{101} < b_{101}$ (C) $s < t$ and $a_{101} > b_{101}$ (D) $s < t$ and $a_{101} < b_{101}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing at the starting line of a race. We have two runners, $a_i$ and $b_i$.

Runner $a_i$ is a steady, disciplined athlete—an Arithmetic Progression (A.P.) who adds the same amount of distance every single second. Runner $b_i$ is a wild, accelerating force—a Geometric Progression (G.P.) who doubles their distance every second.

We are told that at the very start ($i=1$) and at the 51st second ($i=51$), they are exactly neck-and-neck. This is the heart of our problem.

### Unmasking the Sequences

First, let us decode the nature of $b_i$. We are told that $\log_e b_i$ is in an A.P. with a common difference of $\log_e 2$.

Mathematically, this means:
$$\log_e b_{i+1} - \log_e b_i = \log_e 2$$

Using the properties of logarithms, we can rewrite this as $\log_e \left( \frac{b_{i+1}}{b_i} \right) = \log_e 2$. By exponentiating both sides, we reveal the truth:
$$\frac{b_{i+1}}{b_i} = 2$$

Our runner $b_i$ is a G.P. with a common ratio $r=2$. It is an exponential explosion!

### The Geometry of the Sum

Now, consider the sums $s = \sum_{i=1}^{51} a_i$ and $t = \sum_{i=1}^{51} b_i$. We know $a_1 = b_1$ and $a_{51} = b_{51}$.

Because $b_i$ is an exponential function, its graph is strictly convex. In the world of geometry, a straight line (our A.P. runner $a_i$) connecting two points on a convex curve must lie *above* the curve between those two points.

Therefore, for every $i$ between 1 and 51, $a_i > b_i$. It follows naturally that the sum of the A.P. terms must be greater than the sum of the G.P. terms: **$s > t$**.

### The Algebraic Verification

If you prefer the cold, hard logic of algebra, let us calculate the common difference $D$ of our A.P. runner. We know $a_{51} = a_1 + 50D$.

Since $a_{51} = b_{51} = b_1 \cdot 2^{50}$ and $a_1 = b_1$, we find:
$$50D = b_1(2^{50} - 1)$$

When we calculate the sum $s$, we use the classic formula $s = \frac{n}{2}(a_1 + a_n)$, yielding:
$$s = \frac{51}{2} b_1(1 + 2^{50}) = 25.5 b_1(1 + 2^{50})$$

Meanwhile, the sum of our G.P. runner is:
$$t = b_1 \frac{2^{51} - 1}{2 - 1} = b_1(2^{51} - 1)$$

Comparing $s = 25.5 b_1 + 25.5 b_1 \cdot 2^{50}$ with $t = 2 b_1 \cdot 2^{50} - b_1$, it becomes blindingly obvious that $s$ is significantly larger than $t$. The linear runner has accumulated much more 'area' under their path because they started their growth spurt earlier.

### The Final Sprint: $i=101$

But the race doesn't end at 51. We must look at $a_{101}$ and $b_{101}$. For the A.P. runner:
$$a_{101} = a_1 + 100D$$

Substituting our value for $50D$, we get:
$$a_{101} = b_1 + 2b_1(2^{50} - 1) = b_1(2^{51} - 1)$$

Now look at the G.P. runner:
$$b_{101} = b_1 \cdot 2^{100}$$

Comparing $b_1(2^{51} - 1)$ to $b_1 \cdot 2^{100}$, the exponential runner has left the linear runner in the dust. The power of $2^{100}$ is astronomically larger than $2^{51}$.

Thus, **$b_{101} > a_{101}$**. We have successfully navigated the race; exponential growth is the most powerful force in mathematics.

Select Answer:

Visualized Solution

The Sequences $a_{i}$ and $b_{i}$

Unmasking $b_{i}$ as a G.P.

Aligning $a_{i}$ and $b_{i}$

Graphical Comparison of Sums

Common Difference of $a_{i}$

Calculating Sum $s$ (A.P.)

Calculating Sum $t$ (G.P.)

Algebraic Comparison of $s$ and $t$

Extrapolating the A.P. to $i = 101$

Extrapolating the G.P. to $i = 101$

Comparing $a_{101}$ and $b_{101}$

The Sigma Insight: Relation Between A.M., G.M., and H.M.

Analyzing the Setup

Unmasking the Sequences

The Geometry of the Sum

The Algebraic Verification

The Final Sprint: $i = 101$

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Select Answer:

Visualized Solution

The Sequences .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ai​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } bi​

Unmasking .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } bi​ as a G.P.

Aligning .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ai​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } bi​

Graphical Comparison of Sums

Common Difference of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ai​

Calculating Sum .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } s (A.P.)

Calculating Sum .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t (G.P.)

Algebraic Comparison of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } s and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t

Extrapolating the A.P. to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } i=101

Extrapolating the G.P. to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } i=101

Comparing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a101​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b101​

The Sigma Insight: Relation Between A.M., G.M., and H.M.

Analyzing the Setup

Unmasking the Sequences

The Geometry of the Sum

The Algebraic Verification

The Final Sprint: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } i=101

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If x1​,x2​,…,xn​ are any real numbers and n is any positive integer, then

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The Sequences $a_{i}$ and $b_{i}$

Unmasking $b_{i}$ as a G.P.

Aligning $a_{i}$ and $b_{i}$

Common Difference of $a_{i}$

Calculating Sum $s$ (A.P.)

Calculating Sum $t$ (G.P.)

Algebraic Comparison of $s$ and $t$

Extrapolating the A.P. to $i = 101$

Extrapolating the G.P. to $i = 101$

Comparing $a_{101}$ and $b_{101}$

The Final Sprint: $i = 101$

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