Let A₁, A₂, A₃, ... be an increasing geometric progression of positive real numbers. If A₁A₃A₅A₇ = 1

Q: Wait, why is $A_1 A_3 A_5 A_7 = A_4^4$?

This is the beauty of geometric progressions! When you write out the terms, $A_1 = a$, $A_3 = ar^2$, $A_5 = ar^4$, and $A_7 = ar^6$. Multiplying them gives $a^4 r^{12}$. Since $A_4 = ar^3$, then $A_4^4 = (ar^3)^4 = a^4 r^{12}$. The symmetry of the indices around 4 makes this elegant substitution possible.

Q: Can I solve for $a$ and $r$ individually?

Technically, yes, but it is a trap! You would end up with a system of equations that is much harder to solve. In JEE Advanced, always look for the structure of the expression first. By working with $A_4$ and $r^2$ directly, you bypass the need to find $a$ and $r$ separately, saving precious time.

Q: Is there a faster way to find the sum $A_6 + A_8 + A_{10}$?

Yes! Once you have $A_4$ and $r^2$, you can factor out $A_4$ to get $A_4(r^2 + r^4 + r^6)$. Since $r^2 = 6$, this becomes $\frac{1}{6}(6 + 6^2 + 6^3)$. It turns a complex sequence problem into a simple arithmetic calculation.

Question

Let $A_1, A_2, A_3, \dots$ be an increasing geometric progression of positive real numbers. If $A_1 A_3 A_5 A_7 = \frac{1}{1296}$ and $A_2 + A_4 = \frac{7}{36}$, then, the value of $A_6 + A_8 + A_{10}$ is equal to

Options:

(A) 33

(B) 37

(C) 43

(D) 47

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing at the starting line of a race, looking at a sequence of numbers that grows with perfect, rhythmic precision. This is the world of a Geometric Progression (GP).

In this problem, we are dealing with an increasing GP of positive real numbers, $A_1, A_2, A_3, \dots$. The fact that it is increasing implies that the first term $a$ must be positive and the common ratio $r$ must be strictly greater than $1$.

### The Hidden Symmetry in the Product

We are given the product: 
$$A_1 A_3 A_5 A_7 = \frac{1}{1296}$$

If we write these terms in their standard form, $A_n = a r^{n-1}$, we obtain:
$$A_1 = a, \quad A_3 = ar^2, \quad A_5 = ar^4, \quad A_7 = ar^6$$

When we multiply these, we get $a^4 r^{2+4+6} = a^4 r^{12}$. Notice that $A_4 = ar^3$, and raising this to the power of $4$ yields $(ar^3)^4 = a^4 r^{12}$.

Thus, the product simplifies to:
$$A_4^4 = \frac{1}{1296}$$

Since $1296 = 6^4$, we immediately find that $A_4 = \frac{1}{6}$. This value is the key that unlocks the entire problem.

### Bridging to the Common Ratio

Now, we turn to the second condition: $A_2 + A_4 = \frac{7}{36}$. Substituting $A_4 = \frac{1}{6}$, the equation becomes:
$$A_2 + \frac{1}{6} = \frac{7}{36}$$

Subtracting $\frac{1}{6}$ (which is $\frac{6}{36}$) from both sides, we find $A_2 = \frac{1}{36}$. In a GP, the ratio between terms is constant, specifically $\frac{A_4}{A_2} = \frac{ar^3}{ar} = r^2$.

Therefore:
$$r^2 = \frac{1/6}{1/36} = 6$$

We do not need to find $r$ itself, as $r^2 = 6$ is sufficient for the final calculation.

### The Final Calculation

We are asked to find the sum $A_6 + A_8 + A_{10}$. Let us express these in terms of $A_4$ and $r^2$:
$$A_6 = A_4 r^2, \quad A_8 = A_4 r^4, \quad A_{10} = A_4 r^6$$

Factoring out $A_4$, the sum becomes:
$$	ext{Sum} = A_4 (r^2 + r^4 + r^6)$$

Substituting our known values $A_4 = \frac{1}{6}$ and $r^2 = 6$:
$$	ext{Sum} = \frac{1}{6} (6 + 6^2 + 6^3)$$
$$	ext{Sum} = \frac{1}{6} (6 + 36 + 216)$$
$$	ext{Sum} = \frac{1}{6} (258)$$

The final result is:
$$\mathbf{43}$$

Animated Solution for Mathematics - Sequence and Series: Let $A_{1}, A_{2}, A_{3}, \dots$ be an increasing geometric progression of positive real numbers. If $A_{1} A_{3} A_{5} A_{7} = \frac{1}{1296}$ and $A_{2} + A_{4} = \frac{7}{36}$ , then, the value of $A_{6} + A_{8} + A_{10}$ is equal to

Select Answer:

Visualized Solution

Defining the Geometric Progression

Analyzing the Product $A_{1} A_{3} A_{5} A_{7}$

Substituting the Standard Form

Combining the Exponents

Solving for the Fourth Term $A_{4}$

Using the Sum Condition $A_{2} + A_{4}$

Calculating the Second Term $A_{2}$

Finding the Common Ratio Squared $r^{2}$

Setting up the Target Expression

Factoring out $A_{4}$

Substituting Known Values

Final Calculation and Result

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

Analyzing the Setup

The Hidden Symmetry in the Product

Bridging to the Common Ratio

The Final Calculation

Similar Questions

In an increasing geometric progression of positive terms, the sum of the second and sixth terms is $\frac{70}{3}$ and the product of the third and fifth terms is 49. Then the sum of the $4^{th}, 6^{th}$ and $8^{th}$ terms is equal to :

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

Let $a_{1}, a_{2}, a_{3}, \dots$ be a GP of increasing positive numbers. If the product of fourth and sixth terms is 9 and the sum of fifth and seventh terms is 24, then $a_{1} a_{9} + a_{2} a_{4} a_{9} + a_{5} + a_{7}$ is equal to

Let $a_{1}, a_{2}, a_{3}, \dots$ be G.P. of increasing positive terms such that $a_{2} \cdot a_{3} \cdot a_{4} = 64$ and $a_{1} + a_{3} + a_{5} = \frac{813}{7}$ . Then $a_{3} + a_{5} + a_{7}$ is equal to :

Let $a_{1}, a_{2}, a_{3}, \dots$ be a G. P. of increasing positive numbers. If $a_{3} a_{5} = 729$ and $a_{2} + a_{4} = \frac{111}{4}$ , then $24 (a_{1} + a_{2} + a_{3})$ is equal to

If the sum and product of four positive consecutive terms of a G.P., are 126 and 1296, respectively, then the sum of common ratios of all such GPs is

Let the first term $a$ and the common ratio $r$ of a geometric progression be positive integers. If the sum of squares of its first three terms is 33033, then the sum of these three terms is equal to

The sum of first four terms of a geometric progression (G.P.) is $\frac{65}{12}$ and the sum of their respective reciprocals is $\frac{65}{18}$ . If the product of first three terms of the G.P. is 1, and the third term is $α$ , then $2 α$ is

In an increasing, geometric series, the sum of the second and the sixth term is $\frac{25}{2}$ and the product of the third and fifth term is 25. Then, the sum of $4^{t h}, 6^{t h}$ and $8^{t h}$ terms is equal to :

Let $x_{1}, x_{2}, x_{3}, x_{4}$ be in a geometric progression. 2, 7, 9, 5 are subtracted respectively from $x_{1}, x_{2}, x_{3}, x_{4}$ then the resulting numbers are in an arithmetic progression. Then the value of $\frac{1}{24} (x_{1} x_{2} x_{3} x_{4})$ is :