If (S_n=n(6n-1)), find the sum from the (31)st term to the (50)th term.
Answer and explanation
Correct answer: (9580)
Here \(S_n=n(6n-1)\) represents the sum of the first \(n\) terms. A range beginning at the 31st term is found by taking the total through the 50th term and removing the total through the 30th term. This works because \(S_{30}\) contains precisely the terms that come before the required range.
Evaluate \(S_{50}=50(6\cdot50-1)=50(299)=14950\). Next, \(S_{30}=30(6\cdot30-1)=30(179)=5370\). Thus the sum from the 31st through the 50th term is \(S_{50}-S_{30}=14950-5370=9580\). Therefore option D is correct. The number of included terms is 20, from 31 to 50 inclusive, which is consistent with this subtraction method.
Frequently asked questions
What is the correct answer to this question?
(9580)
Why is this the correct answer?
Here \(S_n=n(6n-1)\) represents the sum of the first \(n\) terms. A range beginning at the 31st term is found by taking the total through the 50th term and removing the total through the 30th term. This works because \(S_{30}\) contains precisely the terms that come before the required range.
Evaluate \(S_{50}=50(6\cdot50-1)=50(299)=14950\). Next, \(S_{30}=30(6\cdot30-1)=30(179)=5370\). Thus the sum from the 31st through the 50th term is \(S_{50}-S_{30}=14950-5370=9580\). Therefore option D is correct. The number of included terms is 20, from 31 to 50 inclusive, which is consistent with this subtraction method.
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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