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Example 14 - Sum of first three terms of GP is 13/12, product - Geometric Progression(GP): Formulae based

  1. Chapter 9 Class 11 Sequences and Series
  2. Serial order wise
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Example 14 The sum of first three terms of a G.P. is 13/12 and their product is โ€“ 1. Find the common ratio and the terms. Let the three terms in G.P. be ๐‘Ž/๐‘Ÿ, a, ar here 1st term of G.P. = ๐‘Ž/๐‘Ÿ 2nd term of G.P. = a 3rd term of G.P. = ar It is given that Sum of first three terms of G.P. = 13/12 i.e. a/r + a + ar = 13/12 And product of first three term = -1 (a/r) ร— (a) ร— (ar) = -1 a3 = โ€“ 1 a3 = (-1)3 โˆด a = โ€“ 1 Putting value of a in (1) a/r + a + ar = 13/12 - 1/r + (-1) + (-1)r = 13/12 - 1/r - 1 โ€“ r = 13/12 (โˆ’ 1 โˆ’r โˆ’r2)/r = 13/12 (โˆ’ 1 โˆ’r โˆ’r2)/r = 13/12 12( -1 โ€“ r โ€“ r2 )= 13r -12 โ€“ 12r โ€“ 12r2 = 13r -12 - 12r - 12r2 โ€“ 13r = 0 -12 - 25r - 12r2 = 0 -(12 + 25r + 12r2)= 0 12r2 + 25r + 12= 0 This equation is of the form ax2 + bx + c = 0 where a = 12 b = 25 c = 12 & x = r 12r2 + 25r + 12= 0 where a = 12, b = 25,c = 12, x = r solution of equation is x = (โˆ’๐‘ ยฑ โˆš(๐‘2 โˆ’ 4๐‘Ž๐‘))/29 r = (โˆ’25 ยฑ โˆš((25)^2 โˆ’ 4 ร— 12 ร— 12))/(2 ร— 12) r = (โˆ’25 ยฑ โˆš( 625 โˆ’576))/24 r = (โˆ’25 ยฑ โˆš49)/24 r = (โˆ’25 ยฑ 7)/24 r = (โˆ’25 ยฑ 7)/24 Now we have a = -1 r = - 3/4 & r = - 4/3 Taking r = - 3/4, a = -1 1st term of G.P. = a/r = (โˆ’1)/(โˆ’3/4) = 4/3 2nd term of G.P. = a = -1 3nd term of G.P. = ar = (-1)((โˆ’3)/4) = 3/4 Hence the three term of G.P. are 4/3, -1, 3/4 Taking r = (โˆ’4)/3, a = -1 1st term of G.P. = a/r = (โˆ’1)/((โˆ’4)/3) = 3/4 2nd term of G.P. = a = -1 3nd term of G.P. = ar = (-1)(4/3) =4/3 Hence the three term of G.P. are 3/4, -1, 4/3 Hence first three terms of G.P. are 3/4, -1, 4/3 for r = (โˆ’3)/4 and 3/4, -1, 4/3 for r = (โˆ’4)/3

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.
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