# Misc 8 - Chapter 9 Class 11 Sequences and Series (Term 1)

Last updated at May 29, 2018 by Teachoo

Miscellaneous

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Misc 2

Misc 3 Important

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Misc 5

Misc 6 Important

Misc 7 Important

Misc 8 You are here

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Misc 11

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Misc 21 (i) Important

Misc 21 (ii)

Misc 22 Important

Misc 23 Important

Misc 24 Deleted for CBSE Board 2022 Exams

Misc 25 Important Deleted for CBSE Board 2022 Exams

Misc 26 Deleted for CBSE Board 2022 Exams

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Misc 8 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