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Example 13 - How many terms of GP 3, 3/2, 3/4,... are needed - Geometric Progression(GP): Formulae based

Example 13 - Chapter 9 Class 11 Sequences and Series - Part 2
Example 13 - Chapter 9 Class 11 Sequences and Series - Part 3

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Example 8, How many terms of the G.P. 3, 3/2, 3/4 , , ,... are needed to give the sum 3069/512 ? Here First term = a = 3, Common ratio r = (3/2)/3 = 3/(2 3) = 1/2 We know that sum of n term = ( ( 1 ^ ))/(1 ) Sn = (a(1 ^ ))/(1 r) Given that Sn = 3069/512 & we need to find n. 3069/512 = (a(1 ^ ))/(1 ) 3069/512 = (3[1 (1/2)^ ])/(1 1/2) 3069/512 =(3[1 (1/2)^ ])/( 1/2) 3069/512 = 6[1 (1/2)^ ] 3069/(512 6) = 1 (1/2)^ 3069/3072 = 1 (1/2)^ 1 (1/2)^ = 3069/3072 " " (1/2)^ " = " ("1 " 3069/3072) (1/2)^ = ((3072 3069)/3072) (1/2)^ = (3/3072) (1/2)^ = (1/1024) (1/2)^ = (1/2)10 Comparing powers n = 10 Hence 10 terms are needed to give sum 3069/512

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.