# Ex 9.2, 9

Last updated at March 9, 2017 by Teachoo

Last updated at March 9, 2017 by Teachoo

Transcript

Ex 9.2 , 9 The sums of n terms of two arithmetic progressions are in the ratio 5n + 4: 9n + 6. Find the ratio of their 18th terms. There are two AP with different first term and common difference For the first AP Let first term be = a Common difference = d Sum of n terms = Sn = ๐/2 [2a + (n โ 1)d] & nth term = an = a + (n โ 1)d For the second AP Let first term be = A common difference = D Sum of n terms = Sn = ๐/2 [2A + (n โ 1)D] & nth term = An = A + (n โ 1)D We need to find ratio of their 18th term i.e. (18๐กโ ๐ก๐๐๐ ๐๐ 1๐ ๐ก ๐ด๐ )/(18๐กโ ๐ก๐๐๐ ๐๐ 2๐๐ ๐ด๐) = (๐18 ๐๐ 1๐ ๐ก ๐ด๐ )/(๐ด18 ๐๐ 2๐๐ ๐ด๐) = (a + (18 โ 1)d)/(A + (18 โ 1)D) = (๐ + 17๐)/(A + 17D) is given that (Sum of n terms of first A๐)/(Sum of n terms of second A๐) = (5n+4)/(9n+6) (๐/2[2๐+(๐ โ 1)๐])/((๐ )/2[2๐ด+(๐ โ 1)๐ท]) = (5n+4)/(9n+6) ( [2๐+(๐ โ 1)๐])/( [2๐ด+(๐ โ 1)๐ท]) = (5n+4)/(9n+6) ( 2(a +((๐ โ1)/2)d))/( 2(A +((๐ โ1)/2)D) ) = (5n+4)/(9n+6) ( (a +((๐ โ1)/2)d))/( (A +((๐ โ1)/2)D) ) = (5n+4)/(9n+6) We have to find (๐ + 17๐)/(A + 17D) Hence, (๐ โ1)/2 = 17 n โ 1 = 17 ร 2 n โ 1 = 34 n = 34 + 1 n = 35 Putting n = 35 in (1) ( (a +((35 โ1)/2)d))/( (A +((35 โ1)/2)D) ) " "= (5(35)+4)/(9(35)+6) ( (a +(34/2)d))/( (A +(34/2)D) )= (175 + 4)/(315 + 6) ([a + 17d])/( [A + 17D]) = 179/321 Therefore (18๐กโ ๐ก๐๐๐ ๐๐ 1๐ ๐ก ๐ด๐ )/(18๐กโ ๐ก๐๐๐ ๐๐ 2๐๐ ๐ด๐) = 179/321 Hence the ratio of 18th term of 1st AP and 18th term of 2nd AP is 179 : 321

Example 6
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Ex 9.2, 5 Important

Ex 9.2, 9 Important You are here

Ex 9.2, 15 Important

Ex 9.2, 17 Important

Example 14 Important

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

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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

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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/.