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Last updated at May 29, 2018 by Teachoo

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Ex 9.3, 12 The sum of first three terms of a G.P. is 39/10 and their product is 1. Find the common ratio and the terms. 1st term of G.P = / 2st term of G.P = a 3st term of G.P = ar It is given that sum of first three terms = 39/10 i.e. / + a + ar = 39/10 Also, product of first three terms is 1 / a ar = 1 a3 / = 1 a3 = 1 a3 = (1)3 a = 1 putting a = 1 in (1) / + a + ar = 39/10 1/ + 1 + 1(r) = 39/10 (1+ + 2)/ = 39/10 10(1 + r + r2) = 39r 10(1 + r + r2) = 39r 10 + 10r + 10r2 = 39r 10 + 10r + 10r2 39r = 0 10r2 + 10r 39r + 10 = 0 10r2 29r + 10 = 0 The above equation is of the form ax2 + bx + c = 0 Here a = 10 b = -29 c = 10 & x = r Solutions are x = ( ( 2 4 ))/2 r = ( ( 29) (( 29)2 4 10 10))/(2 10) r = ( ( 29) (( 29)2 4 10 10))/(2 10) r = (29 (841 400))/20 r = (29 441)/20 r = (29 ((21)2))/20 r = (29 21)/20 Taking a = 1 & r = 5/2 First term of G.P = / = 1/(5/2) = 2/5 2nd term of G.P = a = 1 3nd term of G.P = ar =1 5/2 = 5/2 Hence first three terms of G.P are 5/2, 1, 5/2 for r = 5/2 Taking a = 1 & r = 2/5 1st term of G.P = / = 1/(2/5) = 5/2 2nd term of G.P = a = 1 3rd term of G.P = ar = 1 2/5 = 2/5 Hence first three term of G.P are 5/2, 1, 2/5 Hence first three term of G.P are 2/5, 1, 5/2 for r = 5/2 & 5/2, 1, 2/5 for r = 2/5

Ex 9.3

Ex 9.3, 1

Ex 9.3, 2

Ex 9.3, 3 Important

Ex 9.3, 4

Ex 9.3, 5

Ex 9.3, 6

Ex 9.3, 7

Ex 9.3, 8

Ex 9.3, 9

Ex 9.3, 10

Ex 9.3, 11 Important

Ex 9.3, 12 You are here

Ex 9.3, 13

Ex 9.3, 14

Ex 9.3, 15

Ex 9.3, 16

Ex 9.3, 17 Important

Ex 9.3, 18 Important

Ex 9.3, 19

Ex 9.3, 20

Ex 9.3, 21

Ex 9.3, 22 Important

Ex 9.3, 23

Ex 9.3, 24

Ex 9.3, 25

Ex 9.3, 26

Ex 9.3, 27 Important

Ex 9.3, 28 Important

Ex 9.3, 29 Important

Ex 9.3, 30

Ex 9.3, 31

Ex 9.3, 32

Chapter 9 Class 11 Sequences and Series

Serial order wise

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.