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Ex 9.3, 12 - Sum of first three terms of a GP is 39/10 - Geometric Progression(GP): Formulae based

  1. Chapter 9 Class 11 Sequences and Series
  2. Serial order wise
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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‘Ž/๐‘Ÿ 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‘Ž/๐‘Ÿ + a + ar = 39/10 Also, product of first three terms is 1 ๐‘Ž/๐‘Ÿ ร— a ร— ar = 1 a3 ร— ๐‘Ÿ/๐‘Ÿ = 1 a3 = 1 a3 = (1)3 a = 1 putting a = 1 in (1) ๐‘Ž/๐‘Ÿ + a + ar = 39/10 1/๐‘Ÿ + 1 + 1(r) = 39/10 (1+ ๐‘Ÿ +๐‘Ÿ2)/๐‘Ÿ = 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 = (โˆ’๐‘ ยฑ โˆš(๐‘2 โˆ’4๐‘Ž๐‘))/2๐‘Ž r = (โˆ’(โˆ’29) ยฑ โˆš((โˆ’29)2 โˆ’4 ร—10 ร—10))/(2 ร—10) r = (โˆ’(โˆ’29) ยฑ โˆš((โˆ’29)2 โˆ’4 ร—10 ร—10))/(2 ร—10) r = (29 ยฑ โˆš(841 โˆ’ 400))/20 r = (29 ยฑ โˆš441)/20 r = (29 ยฑ โˆš((21)2))/20 r = (29 ยฑ 21)/20 Taking a = 1 & r = 5/2 First term of G.P = ๐‘Ž/๐‘Ÿ = 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‘Ž/๐‘Ÿ = 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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