Example 14 - Chapter 9 Class 11 Sequences and Series

Transcript

Example 14 The sum of first three terms of a G.P. is 13/12 and their product is – 1. Find the common ratio and the terms. Let the three terms in G.P. be 𝑎/𝑟, a, ar here 1st term of G.P. = 𝑎/𝑟 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) × (a) × (ar) = -1 a3 = – 1 a3 = (-1)3 ∴ a = – 1 Putting value of a in (1) a/r + a + ar = 13/12 - 1/r + (-1) + (-1)r = 13/12 - 1/r - 1 – r = 13/12 (− 1 −r −r2)/r = 13/12 (− 1 −r −r2)/r = 13/12 12( -1 – r – r2 )= 13r -12 – 12r – 12r2 = 13r -12 - 12r - 12r2 – 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 = (−𝑏 ± √(𝑏2 − 4𝑎𝑐))/29 r = (−25 ± √((25)^2 − 4 × 12 × 12))/(2 × 12) r = (−25 ± √( 625 −576))/24 r = (−25 ± √49)/24 r = (−25 ± 7)/24 r = (−25 ± 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 = (−1)/(−3/4) = 4/3 2nd term of G.P. = a = -1 3nd term of G.P. = ar = (-1)((−3)/4) = 3/4 Hence the three term of G.P. are 4/3, -1, 3/4 Taking r = (−4)/3, a = -1 1st term of G.P. = a/r = (−1)/((−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 = (−3)/4 and 3/4, -1, 4/3 for r = (−4)/3