Question
Download Solution PDFWhich of the following statement is correct?
I. The value of 1002 - 992 + 982 - 972 + 962 - 952 + 942 - 932 + ...... + 222 - 212 is 4840.
II. The value of
\(\rm \left( {{k^2}+\frac{1}{{{k^2}}}} \right)\left( {k - \frac{1}{k}} \right)\left( {{k^4}+\frac{1}{{{k^4}}}} \right)\left( {k+\frac{1}{k}} \right)\left( {{k^4}-\frac{1}{{{k^4}}}} \right)\) is k16 - \(\frac{1}{k^{16}}\).
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFGiven:
I. The value of 1002 - 992 + 982 - 972 + 962 - 952 + 942 - 932 + ...... + 222 - 212 is 4840.
II. The value of
\(\rm \left( {{k^2}+\frac{1}{{{k^2}}}} \right)\left( {k - \frac{1}{k}} \right)\left( {{k^4}+\frac{1}{{{k^4}}}} \right)\left( {k+\frac{1}{k}} \right)\left( {{k^4}-\frac{1}{{{k^4}}}} \right)\) is k16 - \(\frac{1}{k^{16}}\).
Concept used:
A, (A + D), (A + 2D), ....., L is an arithmetic progression with the first term A and common difference D. (L being the Nth term)
Number of terms, N = (L - A)/D + 1
Sum of all terms = (A + L)/2 × N
Calculation:
II. According to the concept,
1002 - 992 + 982 - 972 + 962 - 952 + 942 - 932 + ...... + 222 - 212
(100 + 99)(100 - 99) + (98 + 97)(98 - 97)+...........(22 + 21)(22 - 21)
So, the series will be
100 + 99 + 98 + 97 .............................................22 + 21.
The first term is 100 and the last is 21, the difference is 1.
Sum = (100 + 21) × 80/2 = 121 × 40 = 4840
II. \(\rm \left( {{k^2}+\frac{1}{{{k^2}}}} \right)\left( {k - \frac{1}{k}} \right)\left( {{k^4}+\frac{1}{{{k^4}}}} \right)\left( {k+\frac{1}{k}} \right)\left( {{k^4}-\frac{1}{{{k^4}}}} \right)\)
⇒ \((k + \frac {1}{k}) (k - \frac {1}{k}) (k^2 + \frac {1}{k^2}) (k^4 + \frac {1}{k^4}) (k^4 - \frac {1}{k^4})\)
⇒ \((k^2 - \frac {1}{k^2}) (k^2 + \frac {1}{k^2}) (k^4 + \frac {1}{k^4}) (k^4 - \frac {1}{k^4})\)
⇒ \((k^4 - \frac {1}{k^4}) (k^4 + \frac {1}{k^4}) (k^4 - \frac {1}{k^4})\)
⇒ \((k^{8} - \frac {1}{k^{8}}) (k^4 - \frac {1}{k^4})\) ≠ \((k^{16} - \frac {1}{k^{16}}) \)
∴ Only I is true.
Last updated on Jul 19, 2025
-> The SSC CGL Notification 2025 has been announced for 14,582 vacancies of various Group B and C posts across central government departments.
-> CSIR NET City Intimation Slip 2025 has been released @csirnet.nta.ac.in.
-> The SSC CGL Tier 1 exam is scheduled to take place from 13th to 30th August 2025 in multiple shifts.
-> Aspirants should visit the official website @ssc.gov.in 2025 regularly for CGL Exam updates and latest announcements.
-> Candidates had filled out the SSC CGL Application Form from 9 June to 5 July, 2025. Now, 20 lakh+ candidates will be writing the SSC CGL 2025 Exam on the scheduled exam date. Download SSC Calendar 2025-25!
-> In the SSC CGL 2025 Notification, vacancies for two new posts, namely, "Section Head" and "Office Superintendent" have been announced.
-> Candidates can refer to the CGL Syllabus for a better understanding of the exam structure and pattern.
-> The CGL Eligibility is a bachelor’s degree in any discipline, with the age limit varying from post to post.
-> The SSC CGL Salary structure varies by post, with entry-level posts starting at Pay Level-4 (Rs. 25,500 to 81,100/-) and going up to Pay Level-7 (Rs. 44,900 to 1,42,400/-).
-> Attempt SSC CGL Free English Mock Test and SSC CGL Current Affairs Mock Test.