# Example 14

Last updated at March 9, 2017 by Teachoo

Last updated at March 9, 2017 by Teachoo

Transcript

Example 14 The sum of first three terms of a G.P. is 13/12 and their product is – 1. Find the common ratio and the terms. Let the three terms in G.P. be 𝑎/𝑟, a, ar here 1st term of G.P. = 𝑎/𝑟 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) × (a) × (ar) = -1 a3 = – 1 a3 = (-1)3 ∴ a = – 1 Putting value of a in (1) a/r + a + ar = 13/12 - 1/r + (-1) + (-1)r = 13/12 - 1/r - 1 – r = 13/12 (− 1 −r −r2)/r = 13/12 (− 1 −r −r2)/r = 13/12 12( -1 – r – r2 )= 13r -12 – 12r – 12r2 = 13r -12 - 12r - 12r2 – 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 = (−𝑏 ± √(𝑏2 − 4𝑎𝑐))/29 r = (−25 ± √((25)^2 − 4 × 12 × 12))/(2 × 12) r = (−25 ± √( 625 −576))/24 r = (−25 ± √49)/24 r = (−25 ± 7)/24 r = (−25 ± 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 = (−1)/(−3/4) = 4/3 2nd term of G.P. = a = -1 3nd term of G.P. = ar = (-1)((−3)/4) = 3/4 Hence the three term of G.P. are 4/3, -1, 3/4 Taking r = (−4)/3, a = -1 1st term of G.P. = a/r = (−1)/((−4)/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 = (−3)/4 and 3/4, -1, 4/3 for r = (−4)/3

Example 6
Important

Ex 9.2, 5 Important

Ex 9.2, 9 Important

Ex 9.2, 15 Important

Ex 9.2, 17 Important

Example 14 Important You are here

Example 15 Important

Ex 9.3, 3 Important

Ex 9.3, 11 Important

Ex 9.3, 17 Important

Ex 9.3, 18 Important

Ex 9.3, 22 Important

Ex 9.3, 28 Important

Ex 9.3, 29 Important

Ex 9.4.4 Important

Ex 9.4, 7 Important

Ex 9.4, 9 Important

Ex 9.4, 10 Important

Example 23 Important

Misc 16 Important

Misc 19 Important

Misc 25 Important

Misc 28 Important

Misc 32 Important

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 7 years. He provides courses for Mathematics and Science from Class 6 to 12. You can learn personally from here https://www.teachoo.com/premium/maths-and-science-classes/.