100 या 100 से कम हरों वाली सभी उचित भिन्नों का योगफल होता है :

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Haryana CET Previous Year Paper (Held On: 6 Nov 2022 Shift 2)
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  1. 2124
  2. 2475
  3. \(22\frac{1}{4}\)
  4. 1925
  5. उत्तर नहीं देना चाहते हैं। 

Answer (Detailed Solution Below)

Option 2 : 2475
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Detailed Solution

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प्रयुक्त सूत्र :

(1 + 2 + 3 + 4 +     ......... + 99) = [ n ( n + 1 ) ] ÷ 2

हल :

हम उन सभी उचित भिन्नों को लिख सकते हैं जिनके हर 100 या 100 से कम हैं:

⇒ 1/2 + (1/3 + 2/3) + (1/4 + 2/4 + 3/4) + ...  (1/100 + 2/100 + .. ..  99/100)

⇒ 1/2 + 3/3 + 6/4 + ..... + 4950/100

⇒ 1/2 + 1 + 3/2 + 2 + ........ + 99/2

इस समीकरण को हम निम्न प्रकार से भी लिख सकते हैं:

⇒ 1/2 + 2/2 + 3/2 + ...... + 99/2

⇒ 1/2 ( 1 + 2 + 3 + ..... + 99 )

⇒  /2 [ ( 99 × 100) ÷ 2 ]

⇒ 99 × 25

⇒ 2475

अतः, सही उत्तर "2475" है।  

Latest Haryana CET Updates

Last updated on Jul 9, 2025

->HSSC CET Exam Date 2025 is 26th and 27th July 2025.

->The Haryana HSSC CET 2025 Exam will be held for two days in 4 shifts.

->Earlier, Haryana CET Group C Notice for EWS Certificate was out. A valid format of EWS Certificate has been given in the Notice.

-> Haryana CET Group C Notification 2025 was out on 26th May 2025.

-> The minimum educational qualification to apply for the Common Eligibility Test is 10+2/equivalent 

-> Candidate applying for CET should not be less than 18 years of age and not more than 42 years.

-> Aspirants must go through the Haryana CET Previous Years’ Papers to understand the need for the exam and prepare for the exam in the right direction.

-> Bihar Police Admit Card 2025 has been released at csbc.bihar.gov.in.

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