# Example 14 - Class 11

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Example 14 The sum of first three terms of a G.P. is 13/12 and their product is โ 1. Find the common ratio and the terms. Let the three terms in G.P. be ๐/๐, a, ar here 1st term of G.P. = ๐/๐ 2nd term of G.P. = a 3rd term of G.P. = ar It is given that Sum of first three terms of G.P. = 13/12 i.e. a/r + a + ar = 13/12 And product of first three term = -1 (a/r) ร (a) ร (ar) = -1 a3 = โ 1 a3 = (-1)3 โด a = โ 1 Putting value of a in (1) a/r + a + ar = 13/12 - 1/r + (-1) + (-1)r = 13/12 - 1/r - 1 โ r = 13/12 (โ 1 โr โr2)/r = 13/12 (โ 1 โr โr2)/r = 13/12 12( -1 โ r โ r2 )= 13r -12 โ 12r โ 12r2 = 13r -12 - 12r - 12r2 โ 13r = 0 -12 - 25r - 12r2 = 0 -(12 + 25r + 12r2)= 0 12r2 + 25r + 12= 0 This equation is of the form ax2 + bx + c = 0 where a = 12 b = 25 c = 12 & x = r 12r2 + 25r + 12= 0 where a = 12, b = 25,c = 12, x = r solution of equation is x = (โ๐ ยฑ โ(๐2 โ 4๐๐))/29 r = (โ25 ยฑ โ((25)^2 โ 4 ร 12 ร 12))/(2 ร 12) r = (โ25 ยฑ โ( 625 โ576))/24 r = (โ25 ยฑ โ49)/24 r = (โ25 ยฑ 7)/24 r = (โ25 ยฑ 7)/24 Now we have a = -1 r = - 3/4 & r = - 4/3 Taking r = - 3/4, a = -1 1st term of G.P. = a/r = (โ1)/(โ3/4) = 4/3 2nd term of G.P. = a = -1 3nd term of G.P. = ar = (-1)((โ3)/4) = 3/4 Hence the three term of G.P. are 4/3, -1, 3/4 Taking r = (โ4)/3, a = -1 1st term of G.P. = a/r = (โ1)/((โ4)/3) = 3/4 2nd term of G.P. = a = -1 3nd term of G.P. = ar = (-1)(4/3) =4/3 Hence the three term of G.P. are 3/4, -1, 4/3 Hence first three terms of G.P. are 3/4, -1, 4/3 for r = (โ3)/4 and 3/4, -1, 4/3 for r = (โ4)/3

Example 1

Example 2

Example 3

Example 4

Example 5

Example 6 Important

Example 7

Example 8

Example 9

Example 10

Example 11

Example 12

Example 13

Example 14 Important You are here

Example 15 Important

Example 16

Example 17 Important

Example 18 Important

Example 19 Important

Example 20

Example 21 Important

Example 22

Example 23 Important

Example 24 Important

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 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.