Chapter 9: Problem 41
Let \(a_{1}, a_{2}, a_{3} \ldots \ldots \ldots . .\) be terms on A.P. If \(\frac{a_{1}+a_{2}+\ldots \ldots \ldots a_{p}}{a_{1}+a_{2}+\ldots \ldots \ldots .+a_{q}}=\frac{p^{2}}{q^{2}}, p \neq q\), then \(\frac{a_{6}}{a_{21}}\) equals \([2006]\) (a) \(\frac{41}{11}\) (b) \(\frac{7}{2}\) (c) \(\frac{2}{7}\) (d) \(\frac{11}{41}\)
Short Answer
Step by step solution
Recognize the Sum Formula for an Arithmetic Progression
Write Sum Expressions for p and q Terms
Set Up the Given Relationship
Simplify the Equation
Solve for the Common Difference Condition
Compute the Ratio of Terms
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
AP sum formula
- "\( n \)" represents the number of terms considered in the sequence.
- "\( a \)" is the starting point of the sequence (first term).
- "\( d \)" is the constant difference added to each successive term.