Misc 2 - Chapter 8 Class 11 Sequences and Series
Last updated at April 16, 2024 by Teachoo
Geometric Progression(GP): Formulae based
Ex 8.2, 6
Example 4
Ex 8.2, 1
Example 5 Important
Ex 8.2, 5 (a)
Ex 8.2, 2
Example 6
Ex 8.2, 4
Ex 8.2, 3 Important
Ex 8.2, 17 Important
Example 7 Important
Ex 8.2, 7 Important
Ex 8.2, 10
Ex 8.2, 9 Important
Ex 8.2, 11 Important
Ex 8.2, 8
Ex 8.2, 19
Ex 8.2, 20
Example 8
Ex 8.2, 13
Ex 8.2, 15
Ex 8.2, 16 Important
Ex 8.2, 21
Ex 8.2, 14 Important
Misc 3
Misc 2 You are here
Example 9 Important
Ex 8.2, 12
Example 10 Important
Ex 8.2, 18 Important
Misc 11 (i) Important
Misc 5
Misc 1 Important
Chapter 8 Class 11 Sequences and Series
Concept wise
- Finding Sequences
- Arithmetic Progression (AP): Formulae based
- Arithmetic Progression (AP): Statement
- Arithmetic Progression (AP): Calculation based/Proofs
- Inserting AP between two numbers
- Arithmetic Mean (AM)
-
Geometric Progression(GP): Formulae based
- Geometric Progression(GP): Statement
- Geometric Progression(GP): Calculation based/Proofs
- Inserting GP between two numbers
- Geometric Mean (GM)
- AM and GM (Arithmetic Mean And Geometric mean)
- AP and GP mix questions
- Finding sum from series
- Finding sum from nth number
Transcript
Misc 2 The sum of some terms of G.P. is 315 whose first term and the common ratio are 5 and 2, respectively. Find the last term and the number of terms. Let a be the first term of GP & r be the common ratio of GP It is given that a = 5 & r = 2 (r > 1) Sum of some term of a GP = 315 Let the sum of n terms of GP = 315 Sn = 315 We know that Sum of n terms of GP = (a( ^ 1))/(r 1) Sn = (a( ^ 1))/(r 1) 315 = (a( ^ 1))/(r 1) Putting r = 2 & a = 5 315 = (5(2n 1))/(2 1) 315 = (5(2n 1))/1 315 = 5 (2n 1) 315/5 = 2n 1 63 = 2n 1 64 = 2n (2)6 = 2n comparing powers n = 6 Hence number of terms is 6 We need to find last term of GP Let l be the last term of GP We know that n term of GP an = arn 1 l = arn 1 Putting a = 5 , r = 2 & n = 6 l = 5(2)6 1 = 5(2)5 = 5(2 2 2 2 2) = 5 64 = 320 Hence, the last term is 320
