Consider two G.P.s 2, 2², 2³, ... and 4, 4², 4³, ... of 60 and n terms respectively. If the geometri

Q: Why can't we just take the arithmetic mean of the exponents?

The geometric mean of a set of numbers is the $N$-th root of their product. If the numbers are $2^{x_1}, 2^{x_2}, \dots, 2^{x_N}$, their product is $2^{\sum x_i}$. The $N$-th root is $(2^{\sum x_i})^{1/N} = 2^{(\sum x_i)/N}$. So, you are actually taking the arithmetic mean of the exponents, not the numbers themselves! It's a beautiful property of logarithms.

Q: How do I handle the base 4 in the second sequence?

Don't panic! The key is to unify the base. Since $4 = 2^2$, you can rewrite $4^k$ as $(2^2)^k = 2^{2k}$. This transforms the second sequence into a power-of-2 sequence, allowing you to add the exponents directly to the first sequence's exponents.

Q: Why do we reject the non-integer solution for $n$?

In the context of sequences, $n$ represents the number of terms. A sequence cannot have 7.125 terms! It must be a discrete, positive integer. Always check the physical constraints of your variables after solving a quadratic equation.

Question

Consider two G.P.s $2, 2^2, 2^3, \dots$ and $4, 4^2, 4^3, \dots$ of 60 and $n$ terms respectively. If the geometric mean of all the $60 + n$ terms is $(2)^{\frac{225}{8}}$, then $\sum_{k=1}^{n} k(n-k)$ is equal to :

Options:

(A) 560

(B) 1540

(C) 1330

(D) 2600

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the path to JEE mastery. Today, we are not just solving a problem; we are peeling back the layers of a beautiful mathematical structure.

Imagine standing before two distinct sequences, like two rivers flowing into a single ocean. The first, $2, 2^2, 2^3, \dots, 2^{60}$, is a steady, predictable stream of powers of two. The second, $4, 4^2, 4^3, \dots, 4^n$, is a more aggressive current, doubling its exponent with every step.

Our goal is to find the geometric mean of this combined flow. The geometric mean of $m$ terms $x_1, x_2, \dots, x_m$ is defined as:

$$GM = \left( \prod_{i=1}^{m} x_i ight)^{1/m}$$

### Phase 1: The Power of Exponents

To find the product of the first sequence, $P_1 = 2^1 \cdot 2^2 \cdot 2^3 \dots 2^{60}$, we utilize the law of exponents: $a^x \cdot a^y = a^{x+y}$.

Thus, $P_1 = 2^{(1+2+3+\dots+60)}$. The sum of the first $k$ integers is given by $\frac{k(k+1)}{2}$.

Applying this, the exponent becomes:

$$\frac{60 	imes 61}{2} = 1830$$

Therefore, $P_1 = 2^{1830}$. This serves as the foundation of our journey.

### Phase 2: The Base 4 Transformation

Now, consider the second sequence: $4, 4^2, 4^3, \dots, 4^n$. Since the base is $4 = 2^2$, we rewrite the sequence as $2^2, 2^4, 2^6, \dots, 2^{2n}$.

The product $P_2$ is $2^{(2+4+6+\dots+2n)}$. Factoring out the $2$, we obtain:

$$P_2 = 2^{2(1+2+3+\dots+n)}$$

Using our summation formula again, the exponent is $2 \cdot \frac{n(n+1)}{2} = n(n+1) = n^2 + n$. Thus, $P_2 = 2^{n^2+n}$.

### Phase 3: The Grand Convergence

With $P_1 = 2^{1830}$ and $P_2 = 2^{n^2+n}$, the total product is $P = P_1 \cdot P_2 = 2^{n^2+n+1830}$. The total number of terms is $m = 60+n$.

The geometric mean is:

$$GM = P^{1/m} = \left( 2^{n^2+n+1830} ight)^{1/(60+n)} = 2^{\frac{n^2+n+1830}{60+n}}$$

We are given that this equals $2^{225/8}$. By equating the exponents, we obtain:

$$\frac{n^2+n+1830}{60+n} = \frac{225}{8}$$

### Phase 4: The Quadratic Resolution

Cross-multiplying gives $8(n^2+n+1830) = 225(60+n)$, which simplifies to:

$$8n^2 + 8n + 14640 = 13500 + 225n$$

Rearranging the terms, we arrive at the quadratic equation:

$$8n^2 - 217n + 1140 = 0$$

Using the quadratic formula, we find the roots $n = 20$ or $n = 7.125$. Since $n$ must be an integer, **$n=20$** is our valid solution.

Finally, we calculate the requested sum $\sum_{k=1}^{20} k(20-k) = 20\sum k - \sum k^2$. Using standard summation formulas:

$$20 \left( \frac{20 	imes 21}{2} ight) - \frac{20 	imes 21 	imes 41}{6} = 4200 - 2870 = 1330$$

You have conquered the problem! The final result is **$1330$**.

Animated Solution for Mathematics - Sequence and Series: Consider two G.P.s $2, 2^{2}, 2^{3}, \dots$ and $4, 4^{2}, 4^{3}, \dots$ of 60 and $n$ terms respectively. If the geometric mean of all the $60 + n$ terms is $(2)^{\frac{225}{8}}$ , then $\sum_{k = 1}^{n} k (n - k)$ is equal to :

Select Answer:

Visualized Solution

Identifying the Two Sequences

Product of the First G.P.

Product of the Second G.P.

Total Product of All Terms

Defining Geometric Mean

Equating the Exponents

Forming the Quadratic Equation

Solving for $n$

Selecting the Valid Integer $n$

Setting up the Summation

Applying Summation Formulas

Final Arithmetic Calculation

The Sigma Insight: Sum of Special Series

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Phase 4

Similar Questions

Let $a_{1}, a_{2}, \dots, a_{10}$ be an $A P$ with common difference $- 3$ and $b_{1}, b_{2}, \dots, b_{10}$ be a $GP$ with common ratio $2$ . Let $c_{k} = a_{k} + b_{k}, k = 1, 2, \dots, 10$ . If $c_{2} = 12$ and $c_{3} = 13$ , then $\sum_{k = 1}^{10} c_{k}$ is equal to .

The sum $\sum_{k = 1}^{20} (1 + 2 + 3 + .... + k)$ is _____________.

Let $α = 1^{2} + 4^{2} + 8^{2} + 1 3^{2} + 1 9^{2} + 2 6^{2} + \dots$ upto 10 terms and $β = \sum_{n = 1}^{10} n^{4}$ . If $4 α - β = 55 k + 40$ , then $k$ is equal to

Let $a_{n}$ be $n^{t h}$ term of the series $5 + 8 + 14 + 23 + 35 + 50 + \dots$ and $S_{n} = \sum_{k = 1}^{n} a_{k}$ . Then $S_{30} - a_{40}$ is equal to

The sum to 20 terms of the series $2 \cdot 2^{2} - 3^{2} + 2 \cdot 4^{2} - 5^{2} + 2 \cdot 6^{2} - \dots$ is equal to __________.

The sum $\sum_{k = 1}^{20} (1 + 2 + 3 + ... + k)$ is

If the sum of the first 15 terms of the series $(3/4)^{3} + (1/2)^{3} + (2/3)^{3} + 3^{3} + (3/4)^{3} + \dots$ is equal to $225 k$ , then $k$ is equal to :

The sum of the following series $1 + 6 + \frac{9 ( 1 ^{2} + 2 ^{2} + 3 ^{2} )}{7} + \frac{12 ( 1 ^{2} + 2 ^{2} + 3 ^{2} + 4 ^{2} )}{9} + \frac{15 ( 1 ^{2} + 2 ^{2} + ... + 5 ^{2} )}{11} + ...$ up to 15 terms, is:

Sum of the first $n$ terms of the series $\frac{1}{2} + \frac{3}{4} + \frac{7}{8} + \frac{15}{16} + \dots$ is equal to

If $a_{n} = \frac{- 2}{4 n ^{2} - 16 n + 15}$ , then $a_{1} + a_{2} + \dots + a_{25}$ is equal to :