श्रृंखला \(3 - 1 + \frac{1}{3} - \frac{1}{9} + \ldots \) का योग किसके बराबर है?

This question was previously asked in
NDA (Held On: 9 Sept 2018) Maths Previous Year paper
View all NDA Papers >
  1. 20/9
  2. 9/20
  3. 9/4
  4. 4/9

Answer (Detailed Solution Below)

Option 3 : 9/4
Free
NDA 01/2025: English Subject Test
5.3 K Users
30 Questions 120 Marks 30 Mins

Detailed Solution

Download Solution PDF

धारणा:

यदि a, ar, ar2, …. एक अनंत GP है तो अनंत गुणोत्तर श्रृंखला का योग निम्न द्वारा दिया जाता है।

अनंत GP का योग = \({s_\infty } = \;\frac{a}{{1\; - \;r}}\); |r| < 1

गणना:

यहां, हमें श्रृंखला \(3 - 1 + \frac{1}{3} - \frac{1}{9} + \ldots \) का योग खोजना होगा

जैसा कि हम देख सकते हैं कि अनुक्रम a = 3 और r = - 1/3 के साथ एक GP है

श्रृंखला का योग = S = a/ (1 - r)

श्रृंखला का योग = \(\frac{3}{{1\; - \;\frac{{ - 1}}{3}}} = \;\frac{3}{{1 + \;\frac{1}{3}}} = \;\frac{3}{{\left( {\frac{4}{3}} \right)}} = \;\frac{9}{4}\) 
Latest NDA Updates

Last updated on May 30, 2025

->UPSC has released UPSC NDA 2 Notification on 28th May 2025 announcing the NDA 2 vacancies.

-> A total of 406 vacancies have been announced for NDA 2 Exam 2025.

->The NDA exam date 2025 has been announced for cycle 2. The written examination will be held on 14th September 2025.

-> Earlier, the UPSC NDA 1 Exam Result has been released on the official website.

-> The selection process for the NDA exam includes a Written Exam and SSB Interview.

-> Candidates who get successful selection under UPSC NDA will get a salary range between Rs. 15,600 to Rs. 39,100. 

-> Candidates must go through the NDA previous year question paper. Attempting the NDA mock test is also essential. 

More Sequences and Series Questions

Get Free Access Now
Hot Links: teen patti casino teen patti 100 bonus teen patti master app