# Example 14

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 6
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Ex 9.2, 5 Important

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Example 14 Important You are here

Example 15 Important

Ex 9.3, 3 Important

Ex 9.3, 11 Important

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Ex 9.3, 22 Important

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Ex 9.4.4 Important

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Example 23 Important

Misc 16 Important

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Class 11

Important Question for exams Class 11

- Chapter 1 Class 11 Sets
- Chapter 2 Class 11 Relations and Functions
- Chapter 3 Class 11 Trigonometric Functions
- Chapter 4 Class 11 Mathematical Induction
- Chapter 5 Class 11 Complex Numbers
- Chapter 6 Class 11 Linear Inequalities
- Chapter 7 Class 11 Permutations and Combinations
- Chapter 8 Class 11 Binomial Theorem
- Chapter 9 Class 11 Sequences and Series
- Chapter 10 Class 11 Straight Lines
- Chapter 11 Class 11 Conic Sections
- Chapter 12 Class 11 Introduction to Three Dimensional Geometry
- Chapter 13 Class 11 Limits and Derivatives
- Chapter 14 Class 11 Mathematical Reasoning
- Chapter 15 Class 11 Statistics
- Chapter 16 Class 11 Probability

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.