JEE Exam > JEE Questions > 113 + 123 + 133 + ... + 203 isa)an odd intege...
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113 + 123 + 133 + ... + 203 is
- a)an odd integer divisible by 5
- b)an even integer
- c)multiple of 10
- d)an odd integer but not a multiple of 5
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
113 + 123 + 133 + ... + 203 isa)an odd integer divisible by 5b)an even...
Explanation:
Series Analysis:
The given series is an arithmetic progression where the common difference is 10. It starts from 113 and ends at 203.
Identifying the Pattern:
To find the sum of the series, we can look for a pattern in the numbers. If we observe, the sum of the first and last terms, 113 + 203, gives 316. Similarly, the sum of the second and second last terms, 123 + 193, also gives 316. This pattern continues for all pairs of terms.
Formula for Sum of Arithmetic Progression:
The sum of an arithmetic series can be calculated using the formula:
Sum = n/2 * (first term + last term)
where n is the number of terms in the series.
Calculating the Sum:
In this case, there are (203-113)/10 + 1 = 10 terms in the series.
Sum = 10/2 * (113 + 203) = 5 * 316 = 1580
Answer Justification:
Since the sum of the series is 1580, which is an odd integer, and it is also divisible by 5, the correct answer is option 'A' - an odd integer divisible by 5.
Series Analysis:
The given series is an arithmetic progression where the common difference is 10. It starts from 113 and ends at 203.
Identifying the Pattern:
To find the sum of the series, we can look for a pattern in the numbers. If we observe, the sum of the first and last terms, 113 + 203, gives 316. Similarly, the sum of the second and second last terms, 123 + 193, also gives 316. This pattern continues for all pairs of terms.
Formula for Sum of Arithmetic Progression:
The sum of an arithmetic series can be calculated using the formula:
Sum = n/2 * (first term + last term)
where n is the number of terms in the series.
Calculating the Sum:
In this case, there are (203-113)/10 + 1 = 10 terms in the series.
Sum = 10/2 * (113 + 203) = 5 * 316 = 1580
Answer Justification:
Since the sum of the series is 1580, which is an odd integer, and it is also divisible by 5, the correct answer is option 'A' - an odd integer divisible by 5.
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Question Description
113 + 123 + 133 + ... + 203 isa)an odd integer divisible by 5b)an even integerc)multiple of 10d)an odd integer but not a multiple of 5Correct answer is option 'A'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about 113 + 123 + 133 + ... + 203 isa)an odd integer divisible by 5b)an even integerc)multiple of 10d)an odd integer but not a multiple of 5Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 113 + 123 + 133 + ... + 203 isa)an odd integer divisible by 5b)an even integerc)multiple of 10d)an odd integer but not a multiple of 5Correct answer is option 'A'. Can you explain this answer?.
113 + 123 + 133 + ... + 203 isa)an odd integer divisible by 5b)an even integerc)multiple of 10d)an odd integer but not a multiple of 5Correct answer is option 'A'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about 113 + 123 + 133 + ... + 203 isa)an odd integer divisible by 5b)an even integerc)multiple of 10d)an odd integer but not a multiple of 5Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 113 + 123 + 133 + ... + 203 isa)an odd integer divisible by 5b)an even integerc)multiple of 10d)an odd integer but not a multiple of 5Correct answer is option 'A'. Can you explain this answer?.
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