Misc 2 - Chapter 8 Class 11 Sequences and Series
Last updated at April 16, 2024 by Teachoo
Miscellaneous
Misc 2 You are here
Misc 3
Misc 4 Important
Misc 5
Misc 6
Misc 7 Important
Misc 8
Misc 9
Misc 10 Important
Misc 11 (i) Important
Misc 11 (ii)
Misc 12 Important
Misc 13
Misc 14 Important
Misc 15 Important
Misc 16
Misc 17 Important
Misc 18 Important
Question 1 Deleted for CBSE Board 2025 Exams
Question 2 Deleted for CBSE Board 2025 Exams
Question 3 Important Deleted for CBSE Board 2025 Exams
Question 4 Deleted for CBSE Board 2025 Exams
Question 5 Deleted for CBSE Board 2025 Exams
Question 6 Important Deleted for CBSE Board 2025 Exams
Question 7 Deleted for CBSE Board 2025 Exams
Question 8 Deleted for CBSE Board 2025 Exams
Question 9 Important Deleted for CBSE Board 2025 Exams
Question 10 Deleted for CBSE Board 2025 Exams
Question 11 Important Deleted for CBSE Board 2025 Exams
Question 12 Deleted for CBSE Board 2025 Exams
Question 13 Important Deleted for CBSE Board 2025 Exams
Question 14 Deleted for CBSE Board 2025 Exams
Miscellaneous
Last updated at April 16, 2024 by Teachoo
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