If (S_n=6n^2-5n), find the sum from the (31)st term to the (50)th term.
Answer and explanation
Correct answer: (9500)
The notation \(S_n=6n^2-5n\) gives the sum of the first \(n\) terms, not the value of the \(n\)th term. To obtain the sum beginning with the 31st term and ending with the 50th term, subtract the sum through the 30th term from the sum through the 50th term. This removes terms 1 to 30 and leaves exactly terms 31 to 50.
Calculate \(S_{50}=6(50)^2-5(50)=15000-250=14750\), and \(S_{30}=6(30)^2-5(30)=5400-150=5250\). Therefore the required range sum is \(S_{50}-S_{30}=14750-5250=9500\). Hence option B is correct. Subtracting \(S_{31}\) instead would wrongly omit the 31st term.
Frequently asked questions
What is the correct answer to this question?
(9500)
Why is this the correct answer?
The notation \(S_n=6n^2-5n\) gives the sum of the first \(n\) terms, not the value of the \(n\)th term. To obtain the sum beginning with the 31st term and ending with the 50th term, subtract the sum through the 30th term from the sum through the 50th term. This removes terms 1 to 30 and leaves exactly terms 31 to 50.
Calculate \(S_{50}=6(50)^2-5(50)=15000-250=14750\), and \(S_{30}=6(30)^2-5(30)=5400-150=5250\). Therefore the required range sum is \(S_{50}-S_{30}=14750-5250=9500\). Hence option B is correct. Subtracting \(S_{31}\) instead would wrongly omit the 31st term.
Which subject and chapter does this question cover?
This is a Class 10 Mathematics question. Chapter: Arithmetic Progressions (AP). Topic: Finding the sum of the first $n$ terms of an AP.
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