Misc 10 - Class 11

Transcript

Misc 10 The sum of three numbers in G.P. is 56. If we subtract 1, 7, 21 from these numbers in that order, we obtain an arithmetic progression. Find the numbers. Let the numbers in G.P. be a, ar, and ar2. It is given that Sum of these number is 56 a + ar + ar2 = 56 ar2 = 56 – a – ar Also, When 1, 7, 21 subtracting from these number respectively, the new numbers are in AP So, (a – 1) , (ar – 7) , (ar2 – 21) are in AP ∴ Common difference is same (ar – 7) – (a – 1) = (ar2 – 21) – (ar – 7) ar – a – 6 = ar2 – ar – 14 ar2 – ar – ar + a – 14 + 6 = 0 ar2 – 2ar + a – 8 = 0 From (1) putting ar2 = 56 – a – ar (56 – a – ar) – 2ar + a – 8 = 0 56 – a – ar – 2ar + a – 8 = 0 -a + a – ar – 2ar + 56 – 8 = 0 0 – 3ar + 48 = 0 -3ar = -48 ar = (−48)/(−3) ar = 16 a = 16/𝑟 Putting a = (16 )/𝑟 in (1) a + ar + ar2 = 56 a(1 + r + r2) = 56 16/𝑟(1 + r + r2) = 56 (16 (1+𝑟+𝑟2))/𝑟 = 56 (16 (1+𝑟+𝑟2))/𝑟 = 56 16(1 + r + r2) = 56r 16 + 16r + 16r2 – 56r = 0 16r2 + 16r – 56r + 16 = 0 8(2r2 – 5r + 2) = 0 (2r2 – 5r + 2) = 0/8 2r2 – 5r + 2 = 0 2r2 – 4r – r + 2 = 0 2r(r – 2) – 1(r – 2) = 0 (2r – 1)(r – 2) = 0 2r – 1 = 0 or r – 2 = 0 2r = 1 or r = 2 r = 1/2 or r = 2 Now, finding numbers Thus the numbers are 8, 16, 32 for r = 1/2 & a =32 & 32, 16, 8 for r = 2 & a = 8