Problem 139
Question
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
Step-by-Step Solution
Verified Answer
The product of the 9 terms is 512; Option (B).
1Step 1: 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.
2Step 2: 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, \( a_5 = ar^4 = 2 \).
3Step 3: Write the Formula for the Product
The product of the first \( n \) terms of a GP is given by \( P_n = (ar)(ar^2)(ar^3) imes ext{...} imes ar^{n-1} = a^n r^{1+2+...+(n-1)} \).
4Step 4: 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 \( \frac{9 \times 8}{2} = 36 \).
5Step 5: 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, \( P_9 = \frac{2^9}{r^{36}} \times r^{36} = 2^9 = 512 \).
6Step 6: Select the Correct Option
Comparing with the given options, the product of the 9 terms is 512, which corresponds to option (B).
Key Concepts
Fifth TermProduct of TermsFirst Nine TermsCommon Ratio
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.
- \( a_5 = ar^4 = 2 \)
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.
- \( a = \frac{2}{r^4} \)
- \( P_9 = (\frac{2}{r^4})^9 \times r^{36} \)
- The \( r \) terms cancel out, giving \( P_9 = 2^9 \)
- Finally, the answer is \( P_9 = 512 \).
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:
- \( a \)
- \( ar \)
- \( ar^2 \)
- ...\( ar^8 \)
Common Ratio
The common ratio \( r \) is integral to understanding a GP. It’s the factor by which any term is multiplied to produce the next term. The entire sequence behavior is driven by this constant multiplier.
- It’s evident in the general formula for terms: \( a_n = ar^{n-1} \).
Other exercises in this chapter
Problem 136
Assertion: For every natural number \(n,(n !)^{3}\) G.M. for \(n\) distinct positive quantities
View solution Problem 138
The value of \(2^{1 / 4} \cdot 4^{1 / 8} \cdot 8^{1 / 16} \ldots \infty\) is: (A) 1 (B) 2 (C) \(3 / 2\) (D) 4
View solution Problem 142
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}
View solution Problem 145
The sum of the first \(\mathrm{n}\) terms of the series \(\mathrm{I}^{2}+2 \cdot 2^{2}\) \(+3^{2}+2 \cdot 4^{2}+5^{2}+2 \cdot 6^{2}+\ldots\) is \(\frac{n(n+1)^{
View solution