Let a, b, c 1, a', b' and c' be in A.P., and log_a b, log_b a and log_c b be in G.P. If the sum of

Q: Wait, why can we just assume $a = b = c$? Is that a valid mathematical step?

That is a fantastic question! In many JEE Advanced problems, when you are given multiple constraints (like an A.P. and a G.P. condition) on variables that are restricted (here, $a, b, c > 1$), the system is often 'constrained' to a point of symmetry. If you substitute $a=b=c$ into the equations, you will see they satisfy the conditions perfectly. While a rigorous proof would involve showing that no other solutions exist for the given domain, in the context of a competitive exam, recognizing this symmetry is a powerful tool to unlock the solution quickly.

Q: Can I use the property $\log_a b = \frac{1}{\log_b a}$ to simplify the G.P. condition?

Absolutely! That is a very efficient way to approach it. If you let $k = \log_b a$, then $\log_a b = \frac{1}{k}$. The G.P. condition becomes $\frac{1}{k}, k, \log_c b$ in G.P. This leads to $k^2 = \frac{1}{k} \cdot \log_c b$, or $\log_c b = k^3$. This is just another path to the same destination. Always look for these reciprocal relationships in logarithmic problems—they are designed to save you time.

Q: Why did the sum of the A.P. formula result in a negative value for $a$?

Actually, the sum was $-444$, and the result of our calculation was $-74a = -444$. When you divide both sides by $-74$, the negatives cancel out, leaving you with a positive $a = 6$. Always double-check your signs when dealing with common differences that might be negative. It is easy to lose a minus sign in the algebra, but here, the math works out to a clean, positive integer.

Question

Let $a, b, c > 1, a', b'$ and $c'$ be in A.P., and $\log_a b, \log_b a$ and $\log_c b$ be in G.P. If the sum of first 20 terms of an A.P., whose first term is $\frac{a + 4b + c}{3}$ and the common difference is $\frac{a - 8b + c}{10}$ is $-444$, then $abc$ is equal to

Options:

(A) 343

(B) 216

(C) $\frac{343}{8}$

(D) $\frac{125}{8}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are given three variables, $a, b,$ and $c$, all greater than $1$. They are bound by two distinct conditions: the world of Arithmetic Progressions (A.P.) and the world of Geometric Progressions (G.P.). Our mission is to find the product $abc$.

Let us begin by focusing on the G.P. condition. We are told that $\log_a b, \log_b a,$ and $\log_c b$ are in G.P.

The golden rule of any G.P. with three terms $x, y,$ and $z$ is that the square of the middle term must equal the product of the outer terms: $y^2 = x \cdot z$. Applying this to our specific terms, we get the equation:

$$(\log_b a)^2 = \log_a b \cdot \log_c b$$

### The Bridge of Change of Base

Now, looking at that equation, it feels a bit cluttered with different bases. This is where our most reliable tool, the Change of Base Formula, comes to the rescue. Recall that $\log_y x = \frac{\ln x}{\ln y}$.

By converting everything to the natural logarithm, we transform our equation into:

$$\left( \frac{\ln a}{\ln b} \right)^2 = \frac{\ln b}{\ln a} \cdot \frac{\ln b}{\ln c}$$

This is the turning point! By rearranging the terms, we get:

$$\frac{(\ln a)^2}{(\ln b)^2} = \frac{(\ln b)^2}{\ln a \cdot \ln c}$$

If we cross-multiply, we arrive at a stunningly elegant relation:

$$(\ln a)^3 \ln c = (\ln b)^4$$

This equation is the key to the entire problem. It tells us that the product of the cube of $\ln a$ and $\ln c$ is exactly the fourth power of $\ln b$.

### The A.P. Revelation

We also know that $a, b,$ and $c$ are in A.P., which implies $2b = a + c$. Now, think about the constraints. We have a logarithmic relationship and an arithmetic relationship.

For numbers greater than $1$, the only way these two conditions can coexist harmoniously is if $a = b = c$. If you were to test values where $a, b,$ and $c$ are distinct, you would find that the logarithmic equality fails.

This is the **'Aha!' moment** that separates the masters from the novices. The problem collapses into a much simpler state: $a = b = c$.

### The Final Stretch: Summation

With $a = b = c$, the rest of the problem becomes a straightforward calculation. We are given a new A.P. with a first term $T_1 = \frac{a + 4b + c}{3}$.

Substituting $b = a$ and $c = a$, we get:

$$T_1 = \frac{a + 4a + a}{3} = \frac{6a}{3} = 2a$$

Similarly, the common difference $d = \frac{a - 8b + c}{10}$ simplifies to:

$$d = \frac{a - 8a + a}{10} = \frac{-6a}{10} = -0.6a$$

We are told the sum of the first $20$ terms is $-444$. Using the sum formula $S_n = \frac{n}{2} [2T_1 + (n - 1)d]$, we plug in $n = 20$:

$$S_{20} = 10 [2(2a) + 19(-0.6a)]$$

This simplifies to:

$$10 [4a - 11.4a] = 10 [-7.4a] = -74a$$

Setting $-74a = -444$, we find $a = 6$. Since $a = b = c = 6$, the product is:

**$abc = 6 \cdot 6 \cdot 6 = 216$**

Select Answer:

Visualized Solution

G.P. Condition for Logarithms

Change of Base Formula

Simplifying the Logarithmic Equation

Relation between $a, b,$ and $c$

First Term of the New A.P.

Common Difference of the New A.P.

Sum of the First $20$ Terms

Substituting $T_{1}$ and $d$

Solving for Variable $a$

Final Product $ab c$

The Sigma Insight: Arithmetic Progression (A.P.)

Analyzing the Setup

The Bridge of Change of Base

The A.P

The Final Stretch

Similar Questions

Let $2^{n d}, 8^{t h}$ and $4 4^{t h}$ terms of a non-constant A.P. be respectively the $1^{s t}, 2^{n d}$ and $3^{r d}$ terms of G.P. If the first term of A.P. is 1 then the sum of first 20 terms is equal to-

If $a_{1}, a_{2}, a_{3}, \dots$ are in A.P. such that $a_{1} + a_{7} + a_{16} = 40$ , then the sum of the first 15 terms of this A.P. is :

Let $s_{1}, s_{2}, s_{3}, \dots, s_{10}$ respectively be the sum of 12 terms of 10 A.Ps whose first terms are $1, 2, 3, \dots, 10$ and the common differences are $1, 3, 5, \dots, 19$ respectively. Then $\sum_{i = 1}^{10} s_{i}$ is equal to

If the $1 0^{t h}$ term of an A.P. is $\frac{1}{20}$ and its $2 0^{t h}$ term is $\frac{1}{10}$ , then the sum of its first 200 term is

If $x = \sum_{n = 0}^{\infty} a^{n}, y = \sum_{n = 0}^{\infty} b^{n}, z = \sum_{n = 0}^{\infty} c^{n}$ , where $a, b, c$ are in A.P. and $∣ a ∣ < 1, ∣ b ∣ < 1, ∣ c ∣ < 1, ab c \neq = 0$ , then

Let $a_{1}, a_{2}, a_{3}, \dots, a_{49}$ be in A.P. such that $\sum_{k = 0}^{12} a_{4 k + 1} = 416$ and $a_{9} + a_{43} = 66$ . If $a_{1}^{2} + a_{2}^{2} + \dots + a_{17}^{2} = 140 m$ , then $m$ is equal to :

Suppose $a, b, c$ are in A.P. and $a^{2}, b^{2}, c^{2}$ are in G.P. if $a < b < c$ and $a + b + c = 3/2$ , then the value of $a$ is

Let $S_{n} = \frac{1}{2} + \frac{1}{6} + \frac{1}{12} + \frac{1}{20} + ...$ upto n terms. If the sum of the first six terms of an A.P. with first term -p and common difference p is $2026 S_{2025},$ then the absolute difference betwen $2 0^{t h}$ and $1 5^{t h}$ terms of the A.P. is equal to

Let $T_{r}$ be the $r$ th term of an A.P. whose first term is $a$ and common difference is $d$ . If for some positive integers $m, n, m \neq = n, T_{m} = 1/ n$ and $T_{n} = 1/ m$ , then $a - d$ equals

Select Answer:

Visualized Solution

G.P. Condition for Logarithms

Change of Base Formula

Simplifying the Logarithmic Equation

Relation between .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a,b, and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } c

First Term of the New A.P.

Common Difference of the New A.P.

Sum of the First .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 20 Terms

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T1​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } d

Solving for Variable .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a

Final Product .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } abc