If (S_n=4n^2+n), find the sum from the (31)st term to the (45)th term.
Answer and explanation
Correct answer: (4515)
The expression \\(S_n=4n^2+n\\) gives the sum of the first \\(n\\) terms of the arithmetic progression. The sum from the 31st term through the 45th term is obtained by subtracting the sum through the 30th term from the sum through the 45th term. Thus the required quantity is \\(S_{45}-S_{30}\\), because this removes terms 1 through 30 and leaves terms 31 through 45.
Calculate \\(S_{45}=4(45)^2+45=4(2025)+45=8145\\). Also, \\(S_{30}=4(30)^2+30=4(900)+30=3630\\). Hence the required sum is \\(8145-3630=4515\\). Therefore option C is correct. The subtraction uses \\(S_{30}\\), not \\(S_{31}\\), because the 31st term must remain included.
Frequently asked questions
What is the correct answer to this question?
(4515)
Why is this the correct answer?
The expression \\(S_n=4n^2+n\\) gives the sum of the first \\(n\\) terms of the arithmetic progression. The sum from the 31st term through the 45th term is obtained by subtracting the sum through the 30th term from the sum through the 45th term. Thus the required quantity is \\(S_{45}-S_{30}\\), because this removes terms 1 through 30 and leaves terms 31 through 45.
Calculate \\(S_{45}=4(45)^2+45=4(2025)+45=8145\\). Also, \\(S_{30}=4(30)^2+30=4(900)+30=3630\\). Hence the required sum is \\(8145-3630=4515\\). Therefore option C is correct. The subtraction uses \\(S_{30}\\), not \\(S_{31}\\), because the 31st term must remain included.
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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