Question
Download Solution PDFGP ची बेरीज शोधा:
4/5 , 4/25 , 4/125 , 4/625 ,... ते n अटी.
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFदिले:
मालिका 4/5 , 4/25 , 4/125 , 4/625 ,... ते n टर्म आहे
वापरलेले सूत्र:
बेरीज = a (1 - r n ) / (1 - r )
जेथे a = प्रथम पद आणि r = सामान्य फरक
गणना:
सामान्य फरक (r) = (4/25) ÷ (4/5)
⇒ (1/5)
वरील फॉर्म्युलामध्ये r चे मूल्य टाकल्यास आपल्याला मिळते:
बेरीज = \(\frac{4}{5}\) (1 - \(\frac{1}{5}\) n ) / 1 - \(\frac{1}{5}\)
⇒ \(\frac{4}{5}\) (1- ( \(\frac{1}{5}\) ) n ) / \(\frac{4}{5}\)
⇒ (1 - ( \(\frac{1}{5}\) ) n )
म्हणून बरोबर उत्तर आहे "(1 - ( \(\frac{1}{5}\) ) n )".
Last updated on Jun 19, 2025
-> The UP Police Sub Inspector 2025 Notification will be released by the end of June 2025 for 4543 vacancies.
-> A total of 35 Lakh applications are expected this year for the UP Police vacancies..
-> The recruitment is also ongoing for 268 vacancies of Sub Inspector (Confidential) under the 2023-24 cycle.
-> The pay Scale for the post ranges from Pay Band 9300 - 34800.
-> Graduates between 21 to 28 years of age are eligible for this post. The selection process includes a written exam, document verification & Physical Standards Test, and computer typing test & stenography test.
-> Assam Police Constable Admit Card 2025 has been released.