For the two positive numbers a, b, if a, b and 1/18 are in a geometric progression, while 1/a, 10 an

Q: Wait, why can I just divide by $b$ in the equation $360b^3 = 18b^2 + b$?

That is a great question! In algebra, dividing by a variable is dangerous if that variable could be zero. However, the problem explicitly states that $a$ and $b$ are positive numbers. Since $b > 0$, we are guaranteed that $b \neq 0$, making the division perfectly safe and mathematically sound.

Q: Why did we reject the negative root in the quadratic formula?

The problem statement defines $a$ and $b$ as positive numbers. When we solved the quadratic equation for $b$, we obtained two potential values. One was positive and one was negative. Since a negative value for $b$ would contradict the initial constraints of the problem, we discard it as an extraneous solution.

Q: Is there a faster way to solve $20 = \frac{1}{a} + \frac{1}{b}$ without substituting?

You could try to manipulate the expression $\frac{a+b}{ab} = 20$, but substitution is the most direct path here. Because $a$ is already expressed in terms of $b^2$ from the G.P. condition, substitution effectively collapses the system into a single-variable problem, which is the standard and most reliable strategy for JEE-level algebra.

Q: Do I need to check if my values of $a$ and $b$ actually satisfy the original sequences?

It is a fantastic habit! If you plug $a = 1/8$ and $b = 1/12$ back into the sequences, you get $1/8, 1/12, 1/18$ (which is a G.P. with common ratio $2/3$) and $8, 10, 12$ (which is an A.P. with common difference $2$). Everything holds perfectly.

Q: What if I made a mistake in the quadratic formula calculation?

The quadratic formula $b = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}$ is your best friend, but it is easy to trip on signs. Always write out $A, B,$ and $C$ clearly before plugging them in. Here, $B = -18$, so $-B$ becomes $+18$. Double-check your discriminant calculation—$324 + 1440 = 1764$, and $\sqrt{1764} = 42$. Precision is the key to success!

Question

For the two positive numbers $a, b$, if $a, b$ and $\frac{1}{18}$ are in a geometric progression, while $\frac{1}{a}, 10$ and $\frac{1}{b}$ are in an arithmetic progression, then, $16a + 12b$ is equal to

Sigma Percentile · Accepted Answer

### The Harmony of Sequences: A Mathematical Journey

Welcome, fellow traveler of the JEE landscape. Today, we are not just solving an algebra problem; we are uncovering a hidden harmony between two fundamental structures: the Geometric Progression (G.P.) and the Arithmetic Progression (A.P.).

Imagine these sequences as two different rhythms in music—one growing by multiplication, the other by addition. Our goal is to find the values of $a$ and $b$ that allow these two rhythms to coexist perfectly.

### Phase 1: Decoding the Geometric Rhythm

We begin with the sequence $a, b, \frac{1}{18}$ in a G.P. In any G.P., the middle term is the geometric mean of its neighbors.

Mathematically, this is expressed as:

$$b^2 = a \cdot \frac{1}{18}$$

By rearranging this, we find a beautiful relationship:

$$a = 18b^2$$

This equation is our first anchor. It tells us that $a$ is inextricably linked to the square of $b$. Keep this in your pocket; we will need it soon.

### Phase 2: The Arithmetic Bridge

Now, let us look at the second sequence: $\frac{1}{a}, 10, \frac{1}{b}$ in an A.P. Here, the rhythm changes.

In an A.P., the middle term is the arithmetic mean of its neighbors, which gives us the elegant relation:

$$2 \cdot 10 = \frac{1}{a} + \frac{1}{b}$$

Simplifying this, we get $20 = \frac{a+b}{ab}$, or more conveniently:

$$20ab = a + b$$

This is our second anchor. We now have two equations and two unknowns—the classic setup for a perfect algebraic resolution.

### Phase 3: The Convergence

Now, the suspense builds. We take our first anchor, $a = 18b^2$, and substitute it into our second equation, $20ab = a + b$.

This leads us to:

$$20(18b^2)b = 18b^2 + b$$

This simplifies to the cubic equation:

$$360b^3 = 18b^2 + b$$

I know what you are thinking: "A cubic equation? That looks intimidating!" But take a breath. Because we know $b > 0$, we can divide the entire equation by $b$ without fear.

This leaves us with the quadratic equation:

$$360b^2 - 18b - 1 = 0$$

### Phase 4: The Final Resolution

Using the quadratic formula $b = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}$, we plug in our values: $A = 360, B = -18, C = -1$.

The discriminant calculation is:

$$\sqrt{(-18)^2 - 4(360)(-1)} = \sqrt{324 + 1440} = \sqrt{1764} = 42$$

Thus, $b = \frac{18 \pm 42}{720}$. Since $b$ must be positive, we ignore the negative root and find:

$$b = \frac{60}{720} = \frac{1}{12}$$

With $b$ in hand, finding $a$ is a simple matter of substitution:

$$a = 18\left(\frac{1}{12}\right)^2 = 18 \cdot \frac{1}{144} = \frac{1}{8}$$

### The Grand Finale

Finally, we evaluate the expression $16a + 12b$. Substituting our hard-won values, we get:

$$16\left(\frac{1}{8}\right) + 12\left(\frac{1}{12}\right) = 2 + 1 = 3$$

Look at that! After all the complexity, the answer collapses into a simple, elegant integer.

**The final result is 3.**

This is the beauty of JEE mathematics—the path may be winding, but the destination is always a testament to the consistency of the laws of nature. You have successfully navigated the interplay of sequences. Keep this confidence, and carry it into your next challenge!

Animated Solution for Mathematics - Sequence and Series: For the two positive numbers $a, b$ , if $a, b$ and $\frac{1}{18}$ are in a geometric progression, while $\frac{1}{a}, 10$ and $\frac{1}{b}$ are in an arithmetic progression, then, $16 a + 12 b$ is equal to

Enter Numerical Value:

Visualized Solution

Problem Setup

Applying G.P. Condition

Applying A.P. Condition

Substituting $a$ into Equation 2

Simplifying the Equation

Solving for $b$

Calculating the Value of $b$

Calculating the Value of $a$

Final Evaluation of $16 a + 12 b$

The Sigma Insight: Geometric Progression (G.P.)

The Harmony of Sequences

Phase 1

Phase 2

Phase 3

Phase 4

The Grand Finale

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