Problem 49 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If the sum of the second, third and fourth terms of a positive term G.P. is 3 and the sum of its sixth, seventh and eighth terms is 243 , then the sum of the first 50 terms of this G.P. is: [Sep. \(05, \mathbf{2 0 2 0}\) (II)] (a) \(\frac{1}{26}\left(3^{49}-1\right)\) (b) \(\frac{1}{26}\left(3^{50}-1\right)\) (c) \(\frac{2}{13}\left(3^{50}-1\right)\) (d) \(\frac{1}{13}\left(3^{50}-1\right)\)

Step-by-Step Solution

Answer

The sum of the first 50 terms is \( \frac{1}{26}(3^{50}-1) \).

Step 1: Understanding the Structure of a G.P.

In a geometric progression (G.P.), any term can be represented by the formula \( a_n = ar^{n-1} \), where \( a \) is the first term and \( r \) is the common ratio. We need to find the value of \( a \) and \( r \) based on the given information.

Step 2: Define the Terms of the G.P.

Let the second, third, and fourth terms of the G.P. be \( ar, ar^2, ar^3 \). According to the problem, their sum is \( ar + ar^2 + ar^3 = 3 \).

Step 3: Equation from the First Condition

The first equation is \( ar(1 + r + r^2) = 3 \). This simplifies to \( ar(1 + r + r^2) = 3 \).

Step 4: Define the Sixth, Seventh, and Eighth Terms

Similarly, the sixth, seventh, and eighth terms are \( ar^5, ar^6, ar^7 \), with their sum being \( ar^5 + ar^6 + ar^7 = 243 \).

Step 5: Equation from the Second Condition

We have the second equation: \( ar^5(1 + r + r^2) = 243 \).

Step 6: Express the Ratio of the Two Conditions

Divide the second equation by the first equation to eliminate \( a \) and the common sum element \( 1 + r + r^2 \):\[ \frac{ar^5(1 + r + r^2)}{ar(1 + r + r^2)} = \frac{243}{3} \]This simplifies to:\[ r^4 = 81 \]so, \( r = 3 \).

Step 7: Determine the First Term

Now substitute \( r = 3 \) into the first equation to find \( a \):\[ ar(1 + r + r^2) = 3 \]\[ 3a(1 + 3 + 9) = 3 \]\[ a = \frac{1}{13} \].

Step 8: Calculate the Sum of the First 50 Terms

The sum of the first \( n \) terms of a G.P. is given by the formula:\[ S_n = a \frac{r^n - 1}{r - 1} \]For \( n = 50 \), \( a = \frac{1}{13} \), and \( r = 3 \), we have:\[ S_{50} = \frac{1}{13} \frac{3^{50} - 1}{2} = \frac{1}{26}(3^{50} - 1) \].

Key Concepts

Common Ratio

In a geometric progression (G.P.), the common ratio is a fundamental component. It dictates how each consecutive term in the sequence relates to the one before it. If you take any term and divide it by the previous one, you’ll end up with the common ratio, often denoted by the symbol \( r \). This consistency in the ratio is what defines a G.P. For example, if you have the terms \( 2, 6, 18 \), dividing each subsequent term by its predecessor as \( \frac{6}{2} \) and \( \frac{18}{6} \), results in a common ratio of 3. It's important to understand this concept because it allows us to generate any term in the G.P. sequence simply by multiplying the first term repeatedly by the common ratio.

• The common ratio can be either positive or negative, affecting the direction or sign of the terms in the sequence.
• It can often reveal how rapidly the terms in a G.P. are increasing or decreasing.

Understanding the common ratio is crucial when solving problems involving G.P., as it helps pinpoint the structure and behavior of the sequence.

Sum of Terms in G.P.

The concept of summing the terms in a geometric progression is integral, especially when looking at finite or infinite series. The formula to calculate the sum depends on the number of terms we consider. For a finite number of terms, say the first \( n \) terms, the sum \( S_n \) can be found using:\[ S_n = a \frac{r^n - 1}{r - 1} \]where:

• \( a \) is the first term,
• \( r \) is the common ratio,
• \( n \) is the number of terms.

This formula arises from recognizing the repeated multiplication inherent in a G.P. and adjusting for the number of terms involved. The formula captures the compounding effect of each new term being a multiple of the one before it.
In an exercise, knowing how to compute the sum allows you to handle quantities over ranges of terms, offering a concise way to sum them up without having to add each term individually.

First Term in G.P.

Determining the first term of a geometric progression is key to understanding the sequence. The first term, represented by \( a \), is the starting point of the series from which all other terms are derived by multiplying with the common ratio \( r \). It is crucial because, in conjunction with the common ratio, it sets the scale and specific sequence of the progression.
In our specific problem, we found \( a \) by using the conditions provided in the problem. By substituting the known value of the common ratio \( r \), into the equations that relate to the sums of specific terms, we calculated that \( a = \frac{1}{13} \). This process highlights how the first term can often be unearthed through algebraic manipulation and solving simultaneous equations that relate to specific conditions spelled out in G.P. problems.

Formula for Sum of G.P.

The formula for summing a geometric progression is exceptionally handy, as it lets you find the sum of a series of terms without needing to individually compute each term. It’s especially useful for large sequences due to its reliance on just a few variables:- \( a \) for the first term,- \( r \) for the common ratio, and- \( n \) for the total number of terms you wish to sum.
The formula:\[ S_n = a \frac{r^n - 1}{r - 1} \]provides a precise sum for the first \( n \) terms.

• When \( |r| < 1 \), the series converges, giving us a formula for an infinite sum.
• In finite series, which our problem deals with, the term \( r^n \) highlights the exponential growth of the sum as \( n \) increases if \( r > 1 \).

This approach permits efficient solution of complex problems by simplifying the computation of large sequences, helping us with exercises like finding the sum of the first 50 terms in a G.P.