If \(S_n-S_{n-1}=126\) and \(S_m=1953\), what is \(n-m\), where \(S_k=1+2+\cdots+k\)?
Answer and explanation
Correct answer: 64
Use the consecutive-sum identity \(S_n-S_{n-1}=n\). The first condition therefore gives \(n=126\). For the second condition, solve \(S_m=\frac{m(m+1)}2=1953\). Since \(62\cdot63/2=31\cdot63=1953\), we obtain \(m=62\). Hence \(n-m=126-62=64\), so option C is correct. The positive index is selected because a sum of the first natural numbers uses a nonnegative natural index in this school context. Options A, B, and D would correspond to incorrect identification of either the consecutive difference or the index of 1953.
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What is the correct answer to this question?
64
Why is this the correct answer?
Use the consecutive-sum identity \(S_n-S_{n-1}=n\). The first condition therefore gives \(n=126\). For the second condition, solve \(S_m=\frac{m(m+1)}2=1953\). Since \(62\cdot63/2=31\cdot63=1953\), we obtain \(m=62\). Hence \(n-m=126-62=64\), so option C is correct. The positive index is selected because a sum of the first natural numbers uses a nonnegative natural index in this school context. Options A, B, and D would correspond to incorrect identification of either the consecutive difference or the index of 1953.
Which subject and chapter does this question cover?
This is a Class 9 Mathematics question. Chapter: Sequences and Progressions. Topic: Sum of first n natural numbers.
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