Last updated at May 29, 2018 by Teachoo

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Ex 9.3, 12 The sum of first three terms of a G.P. is 39/10 and their product is 1. Find the common ratio and the terms. 1st term of G.P = ๐/๐ 2st term of G.P = a 3st term of G.P = ar It is given that sum of first three terms = 39/10 i.e. ๐/๐ + a + ar = 39/10 Also, product of first three terms is 1 ๐/๐ ร a ร ar = 1 a3 ร ๐/๐ = 1 a3 = 1 a3 = (1)3 a = 1 putting a = 1 in (1) ๐/๐ + a + ar = 39/10 1/๐ + 1 + 1(r) = 39/10 (1+ ๐ +๐2)/๐ = 39/10 10(1 + r + r2) = 39r 10(1 + r + r2) = 39r 10 + 10r + 10r2 = 39r 10 + 10r + 10r2 โ 39r = 0 10r2 + 10r โ 39r + 10 = 0 10r2 โ 29r + 10 = 0 The above equation is of the form ax2 + bx + c = 0 Here a = 10 b = -29 c = 10 & x = r Solutions are x = (โ๐ ยฑ โ(๐2 โ4๐๐))/2๐ r = (โ(โ29) ยฑ โ((โ29)2 โ4 ร10 ร10))/(2 ร10) r = (โ(โ29) ยฑ โ((โ29)2 โ4 ร10 ร10))/(2 ร10) r = (29 ยฑ โ(841 โ 400))/20 r = (29 ยฑ โ441)/20 r = (29 ยฑ โ((21)2))/20 r = (29 ยฑ 21)/20 Taking a = 1 & r = 5/2 First term of G.P = ๐/๐ = 1/(5/2) = 2/5 2nd term of G.P = a = 1 3nd term of G.P = ar =1 ร 5/2 = 5/2 Hence first three terms of G.P are 5/2, 1, 5/2 for r = 5/2 Taking a = 1 & r = 2/5 1st term of G.P = ๐/๐ = 1/(2/5) = 5/2 2nd term of G.P = a = 1 3rd term of G.P = ar = 1 ร 2/5 = 2/5 Hence first three term of G.P are 5/2, 1, 2/5 Hence first three term of G.P are 2/5, 1, 5/2 for r = 5/2 & 5/2, 1, 2/5 for r = 2/5

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Chapter 9 Class 11 Sequences and Series

Serial order wise

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.