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Chapter 10: Problem 139

Fifth term of a GP is 2, then the product of its 9 terms is : \(\quad[2002]\) (A) 256 (B) 512 (C) 1024 (D) None of these

Short Answer

Expert verified
The product of the 9 terms is 512; Option (B).

Step by step solution

01

Understand the Problem

The problem provides that the fifth term of a geometric progression (GP) is 2. We need to find the product of the first 9 terms of this GP.
02

Define the General Term of GP

In a geometric progression, the general term can be given as: \( a_n = ar^{n-1} \), where \( a \) is the first term and \( r \) is the common ratio. Here, \( a_5 = ar^4 = 2 \).
03

Write the Formula for the Product

The product of the first \( n \) terms of a GP is given by \( P_n = (ar)(ar^2)(ar^3) imes ext{...} imes ar^{n-1} = a^n r^{1+2+...+(n-1)} \).
04

Define the Sum of Indices for the Product

The expression \( 1+2+...+(n-1) \) is the sum of the first \( n-1 \) natural numbers, which is \( \frac{(n-1)n}{2} \). For \( n=9 \), this becomes \( \frac{9 \times 8}{2} = 36 \).
05

Calculate the Product of First 9 Terms

Using the expression for the product: \( P_9 = a^9 r^{36} \). We know \( ar^4 = 2 \), or equivalently \( a = \frac{2}{r^4} \). So, \( a^9 = \left(\frac{2}{r^4}\right)^9 = \frac{2^9}{r^{36}} \). Then, \( P_9 = \frac{2^9}{r^{36}} \times r^{36} = 2^9 = 512 \).
06

Select the Correct Option

Comparing with the given options, the product of the 9 terms is 512, which corresponds to option (B).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Fifth Term
In a geometric progression (GP), terms are defined using a specific pattern: each term is the previous term multiplied by a constant called the common ratio, denoted as \( r \). The fifth term in this sequence is given by the formula for the general term:
  • \( a_n = ar^{n-1} \), where \( a \) is the first term of the GP and \( n \) is the position of the term.
In our exercise, the fifth term is 2, which can be represented mathematically as:
  • \( a_5 = ar^4 = 2 \)
Knowing the fifth term helps in finding other quantities related to the sequence. By using this formula and knowing one term, you can backtrack to find the first term or the common ratio if additional information is available.
Product of Terms
The product of terms in a GP is an important concept for problems involving multiple terms. For a sequence of terms in a GP, the product of the first \( n \) terms can be geometrically represented and calculated.
  • The formula is \( P_n = a^n r^{1+2+...+(n-1)} \), simplifying the multiplication of terms.
In our problem, we were tasked to find the product of the first nine terms. Using \( ar^4 = 2 \) (from the fifth term), we express \( a \) in terms of \( r \), and substitute back into our product formula:
  • \( a = \frac{2}{r^4} \)
  • \( P_9 = (\frac{2}{r^4})^9 \times r^{36} \)
  • The \( r \) terms cancel out, giving \( P_9 = 2^9 \)
  • Finally, the answer is \( P_9 = 512 \).
Breaking it down helps in understanding how multiplying terms of a GP works.
First Nine Terms
The first nine terms of a GP involve applying the formula for each term successively, starting from the first term and moving up. These terms, calculated using a combination of the first term \( a \) and the common ratio \( r \), follow a predictable pattern:
  • \( a \)
  • \( ar \)
  • \( ar^2 \)
  • ...\( ar^8 \)
For the entire exercise, the key to finding the product of these nine terms was to streamline calculations through formula use, but you can manually find each term and multiply if needed. Knowing any specific term, like the fifth term, lets you derive the rest with ease, provided you have the common ratio.
Common Ratio
The common ratio \( r \) is integral to understanding a GP. It鈥檚 the factor by which any term is multiplied to produce the next term. The entire sequence behavior is driven by this constant multiplier.
  • It鈥檚 evident in the general formula for terms: \( a_n = ar^{n-1} \).
In practical scenarios, once you have \( r \) and one of the terms (like the fifth), you can find any other term, including the first. For example, if \( ar^4 = 2 \), rearranging gives you key insights into \( a \) when \( r \) is known. This relationship makes understanding and calculating terms of a GP clear and methodical.

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Most popular questions from this chapter

Let \(\alpha, \beta, \gamma\) be the roots of the equation \(3 x^{3}-x^{2}-3 x+1=0 .\) If \(\alpha, \beta, \gamma\) are in H.P. then \(|\alpha-\gamma|=\) (A) \(\frac{1}{3}\) (B) \(\frac{2}{3}\) (C) \(\frac{4}{3}\) (D) None of these

If \(f: R \rightarrow R\) satisfies \(f(x+y)=f(x)+f(y)\), for all \(x, y\) \(\in R\) and \(f(1)=7\), then \(\sum_{r=1}^{n} f(r)\) is \(|2003|\) (A) \(\frac{7 n}{2}\) (B) \(\frac{7(n+1)}{2}\) (C) \(7 n(n+1)\) (D) \(\frac{7 n(n+1)}{2}\)

The consecutive numbers of a three digit number form a G.P. If we subtract 792 from this number, we get a number consisting of the same digits written in the reverse order and if we increase the second digit of the required number by 2, the resulting number forms an A.P. The number is (A) 139 (B) 193 (C) 931 (D) None of these

The sum to \(n\) terms of the series \(\frac{1}{3}+\frac{5}{9}+\frac{19}{27}+\frac{65}{81}+\ldots\) is (A) \(n-\frac{\left(3^{n}-2^{n}\right)}{2^{n}}\) (B) \(n-\frac{2\left(3^{n}-2^{n}\right)}{3^{n}}\) (C) \(2^{n}-1\) (D) \(3^{n}-1\)

If \(a+b+c=3\) and \(a>0, b>0, c>0\), then the greatest value of \(a^{2} b^{3} c^{2}\) is (A) \(\frac{3^{10} \cdot 2^{4}}{7^{7}}\) (B) \(\frac{3^{9} \cdot 2^{4}}{7^{7}}\) (C) \(\frac{3^{8} \cdot 2^{4}}{7^{7}}\) (D) None of these
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