Problem 143 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

In a G.P. sum of \(n\) terms is 364 , first term is 1 and the common ratio is 3 . Find \(n\).

Step-by-Step Solution

Answer

The value of \(n\) in the given geometric progression with sum 364, first term 1, and common ratio 3 is 6.

Step 1: Recall the formula for the sum of a G.P.

The formula for the sum of the first \(n\) terms of a geometric progression with first term \(a\) and common ratio \(r\) is given by: \[S_n = \frac{a(r^n - 1)}{r - 1}\] In this G.P, we are given \(S_n = 364\), \(a = 1\), and \(r = 3\). Now, we need to find the value of \(n\).

Step 2: Substitute the given values into the formula

Substitute the given values of \(S_n\), \(a\), and \(r\) into the formula for the sum of a G.P: \[364 = \frac{1(3^n - 1)}{3 - 1}\]

Step 3: Solve the equation for n

Now, we need to solve the equation to find the value of \(n\). First, let's simplify and isolate \(3^n\): \[364 = \frac{3^n - 1}{2}\] Multiply both sides of the equation by 2: \[728 = 3^n - 1\] Now add 1 to both sides: \[729 = 3^n\] Since \(729 = 3^6\), we can infer that \(n = 6\). #Conclusion# The value of \(n\) in the given geometric progression with sum 364, first term 1, and common ratio 3 is 6.

Key Concepts

Sum of Geometric Series

Understanding the sum of a geometric series is essential when dealing with various mathematical problems. It refers to the total accumulated sum when a sequence of numbers (where each number after the first is the product of the previous one and a constant called the common ratio) is added up. The formula to calculate this sum is \( S_n = \frac{a(r^n - 1)}{r - 1} \). Here:

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

This formula helps determine how large the sum will be based on how many terms \(n\) you use from the series. It assumes that the common ratio \(r\) is not equal to 1, which would otherwise make the series not geometric. The formula can simplify complex problems into manageable calculations. Practice using it with different values for \(a\), \(r\), and \(n\) to get comfortable with its application.

First Term

The first term of a geometric series, represented by \(a\), is the initial number in the sequence of numbers being considered. In our problem, the first term is given as 1. Understanding the importance of the first term is crucial because it sets the pattern for the entire sequence. Every term that follows is derived by multiplying this first term by the common ratio, \(r\).The first term plays a significant role in calculating the sum of the series and in identifying the sequence itself. If this term changes, the whole series and its properties will change. This concept emphasizes how foundational the first term is to setting the parameters of the geometric progression.

Common Ratio

The common ratio in a geometric progression is a crucial element that defines how the series develops. It's the factor by which each term is multiplied to get the next term in the series. In our problem, the common ratio \(r\) is 3.Understanding the common ratio helps you predict how fast the sequence grows or shrinks. If the common ratio is greater than 1, the series grows exponentially, while if it's between 0 and 1, it diminishes. A negative common ratio will cause the term signs to alternate. The common ratio is essential for using the sum of geometric series formula effectively, allowing you to determine the growth pattern and overall behavior of the series.

Solving Exponential Equations

Solving exponential equations often comes up when working with geometric series, especially when trying to find the number of terms \(n\). In our problem, simplifying \(364 = \frac{3^n - 1}{2}\) by isolating \(3^n\) involves algebraic manipulation.First, you multiply both sides by 2 to get rid of the fraction. This leaves you with \(728 = 3^n - 1\). Adding 1 to both sides yields \(729 = 3^n\). Recognizing that \(3^6\) equals 729, you determine that \(n = 6\).Approaching exponential equations involves logical reasoning and sometimes recognizing numbers or patterns (like powers of 3 in this case) can significantly simplify the solution process. This practice is not only useful in theoretical mathematics but also in real-life applications where exponential growth or decay is modeled.