If in a G.P. of 64 terms, the sum of all the terms is 7 times the sum of the odd terms of the G.P, t

Q: Wait, why can we just cancel $a(r^{64} - 1)$ from both sides? What if it equals zero?

That is a sharp observation! In mathematics, we must always be careful about dividing by zero. However, in the context of a G.P. problem like this, if $a = 0$, the sequence is trivial (all zeros), and if $r^n = 1$ (where $r \neq 1$), the sum would be zero, which contradicts the problem's premise of a meaningful ratio. We assume a non-trivial G.P. where $a \neq 0$ and $r \neq 1$, allowing us to safely divide by these non-zero terms.

Q: How do we know the odd terms form a G.P. with a common ratio of $r^2$?

Let's look at the indices. The odd terms are $a_1, a_3, a_5, \dots$. In a G.P., the $n$-th term is $a_n = ar^{n-1}$. So, $a_1 = a$, $a_3 = ar^2$, and $a_5 = ar^4$. If you divide any term by the previous one in this new sequence (e.g., $\frac{a_3}{a_1} = \frac{ar^2}{a} = r^2$ or $\frac{a_5}{a_3} = \frac{ar^4}{ar^2} = r^2$), you get a constant value of $r^2$. Since the ratio is constant, it is indeed a G.P.!

Q: Why is the number of terms in the odd sequence exactly $32$?

Think about the total number of terms, which is $64$. The sequence is $a_1, a_2, a_3, \dots, a_{64}$. The odd terms are at positions $1, 3, 5, \dots, 63$. This is an arithmetic progression of indices: $1, 3, 5, \dots, 63$. The number of terms $n$ in an A.P. is given by $\frac{\text{Last} - \text{First}}{\text{Difference}} + 1$. Here, $\frac{63 - 1}{2} + 1 = \frac{62}{2} + 1 = 31 + 1 = 32$. Exactly half of the $64$ terms are odd-positioned!

Question

If in a G.P. of 64 terms, the sum of all the terms is 7 times the sum of the odd terms of the G.P, then the common ratio of the G.P. is equal to

Options:

(A) 7

(B) 4

(C) 5

(D) 6

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the path of mathematics! Today, we are going to unravel a problem that might look like a daunting algebraic mess at first glance, but beneath the surface, it is a beautiful display of symmetry and cancellation.

We are dealing with a Geometric Progression (G.P.) of $64$ terms. Let's define our universe: the first term is $a$, and the common ratio is $r$. Our sequence is $a, ar, ar^2, \dots, ar^{63}$.

### The Total Sum

Before we dive into the specific condition given, let's establish our baseline. The sum of the first $n$ terms of a G.P. is given by the classic formula:

$$S_n = \frac{a(r^n - 1)}{r - 1}$$

For our sequence of $64$ terms, we simply plug in $n = 64$. Thus, the total sum $S_{64}$ is:

$$S_{64} = \frac{a(r^{64} - 1)}{r - 1}$$

Keep this expression in your mind; it is the foundation of our entire journey.

### The Sub-Universe of Odd Terms

The problem introduces a twist: the sum of the odd-positioned terms. Let's list them out to see the pattern: $a_1, a_3, a_5, \dots, a_{63}$.

Substituting our G.P. definition, these are $a, ar^2, ar^4, \dots, ar^{62}$. Look closely at this sequence; it is, itself, a Geometric Progression.

The first term is $a$, and the common ratio for this sub-sequence is $r^2$. Since we are taking every other term from a set of $64$, we have exactly $32$ terms. The sum of these odd terms, $S_{odd}$, is:

$$S_{odd} = \frac{a((r^2)^{32} - 1)}{r^2 - 1} = \frac{a(r^{64} - 1)}{r^2 - 1}$$

### The Algebraic Dance

Now, the problem gives us the key: the total sum is $7$ times the sum of the odd terms. Mathematically, this is $S_{64} = 7 	imes S_{odd}$.

Let's set up the equation:

$$ \frac{a(r^{64} - 1)}{r - 1} = 7 	imes \frac{a(r^{64} - 1)}{r^2 - 1} $$

I know what you are thinking—this looks complicated. But look at the numerator on both sides: $a(r^{64} - 1)$. It is identical!

As long as $a 
eq 0$ and $r 
eq 1$, we can cancel this term entirely. This is the "aha!" moment where the complexity vanishes. We are left with:

$$ \frac{1}{r - 1} = \frac{7}{r^2 - 1} $$

### The Final Resolution

We are almost at the finish line. The denominator on the right, $r^2 - 1$, is a classic difference of squares. We can factor it as $(r - 1)(r + 1)$.

Our equation becomes:

$$ \frac{1}{r - 1} = \frac{7}{(r - 1)(r + 1)} $$

We can cancel the $(r - 1)$ term from both denominators, leaving us with the beautifully simple relation:

$$ 1 = \frac{7}{r + 1} $$

Cross-multiplying gives us $r + 1 = 7$, which leads us directly to the final result:

**$r = 6$**

Isn't it satisfying? What started as a complex relationship between sums of powers collapsed into a simple linear equation. This is the beauty of algebra—when you trust the process and look for the underlying structure, the most intimidating problems often reveal their simplest truths.

Animated Solution for Mathematics - Sequence and Series: If in a G.P. of 64 terms, the sum of all the terms is 7 times the sum of the odd terms of the G.P, then the common ratio of the G.P. is equal to

Select Answer:

Visualized Solution

Defining the G.P. with $64$ Terms

Calculating Total Sum $S_{64}$

Identifying the Odd Terms

Properties of the Odd Terms G.P.

Sum of Odd Terms $S_{o dd}$

Applying the Given Condition

Simplifying the Equation

Using Algebraic Identity

Final Value of $r$

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

Analyzing the Setup

The Total Sum

The Sub-Universe of Odd Terms

The Algebraic Dance

The Final Resolution

Similar Questions

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

If the 2nd, 5th and 9th terms of a non-constant A.P. are in G.P., then the common ratio of this G.P. is:

In a geometric progression consisting of positive terms, each term equals the sum of the next two terms. Then the common ratio of its progression is equals

The $4^{t h}$ term of GP is 500 and its common ratio is $\frac{1}{m}, m \in N$ . Let $S_{n}$ denote the sum of the first $n$ terms of this GP. If $S_{6} > S_{5} + 1$ and $S_{7} < S_{6} + \frac{1}{2}$ , then the number of possible values of $m$ is

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 :

Sum of infinite number of terms of GP is $20$ and sum of their square is $100$ . The common ratio of GP is

Let $a, b$ and $c$ be the 7th, 11th and 13th terms respectively of a non-constant A.P. If these are also the three consecutive terms of a G.P., then $a / c$ 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

Let $a, b$ and $c$ be in G. P. with common ratio $r$ , where $a \neq = 0$ and $0 < r \leq 1/2$ . If $3 a, 7 b$ and $15 c$ are the first three terms of an A. P., then the 4th term of this A. P. is :