Example 8 - Chapter 8 Class 11 Sequences and Series
Last updated at April 16, 2024 by Teachoo
Geometric Progression(GP): Formulae based
Example 4
Ex 8.2, 1
Example 5 Important
Ex 8.2, 5 (a)
Ex 8.2, 2
Example 6
Ex 8.2, 4
Ex 8.2, 3 Important
Ex 8.2, 17 Important
Example 7 Important
Ex 8.2, 7 Important
Ex 8.2, 10
Ex 8.2, 9 Important
Ex 8.2, 11 Important
Ex 8.2, 8
Ex 8.2, 19
Ex 8.2, 20
Example 8 You are here
Ex 8.2, 13
Ex 8.2, 15
Ex 8.2, 16 Important
Ex 8.2, 21
Ex 8.2, 14 Important
Misc 3
Misc 2
Example 9 Important
Ex 8.2, 12
Example 10 Important
Ex 8.2, 18 Important
Misc 11 (i) Important
Misc 5
Misc 1 Important
Geometric Progression(GP): Formulae based
Last updated at April 16, 2024 by Teachoo
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