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Class 10 > Divide 20 into 4 parts which are in A.P. and such that...
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Divide 20 into 4 parts which are in A.P. and such that the product of first and fourth is to product of the second and third in ratio 2:3

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Ans- 2,4,6,8 and 8,6,4,2 Let the four terms be a-3d,a-d,a+d,a+3d.. Then sum of all this will be 20 which will give 4a=20 or a=5 Following second part of the question (a-3d)(a+3d)/(a-d)(a+d) => a^2-9d^2 /a^2-d^2=2/3 25-9d^2 /25-d^2= 2/3 75- 27d^2=50- 2d^2 25=25d^2 D=+1 or d=-1 So sequence...
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Ans- 2,4,6,8 and 8,6,4,2 Let the four terms be a-3d,a-d,a+d,a+3d.. Then sum of all this will be 20 which will give 4a=20 or a=5 Following second part of the question (a-3d)(a+3d)/(a-d)(a+d) => a^2-9d^2 /a^2-d^2=2/3 25-9d^2 /25-d^2= 2/3 75- 27d^2=50- 2d^2 25=25d^2 D=+1 or d=-1 So sequence will be With d=+1 2,4,6,8 And with D=-1 8,6,4,2 read less
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Let the four terms of the AP = a -3d, a - d, a+d and a +3d. a -3d + a - d + a + d +a +3d = 20 4a = 20 a = 5 (a - 3d)(a +3d)/(a - d)(a + d) = 2/3 a2 - 9d2/a2 - d2 = 2/3 3(a2 - 9d2) = 2(a2 - d2) 3a2 - 27d2 = 2a2 - 2d2 a2 = 25d2 we have a = 5, substituting, we get 25 = 25d2 d2...
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Let the four terms of the AP = a -3d, a - d, a+d and a +3d. a -3d + a - d + a + d +a +3d = 20 4a = 20 a = 5 (a - 3d)(a +3d)/(a - d)(a + d) = 2/3 a2 - 9d2/a2 - d2 = 2/3 3(a2 - 9d2) = 2(a2 - d2) 3a2 - 27d2 = 2a2 - 2d2 a2 = 25d2 we have a = 5, substituting, we get 25 = 25d2 d2 = 25/25 d = 1 The four terms are a - 3d = 5 - 3 = 2, a - d = 5 - 1 = 4, a + d = 5 + 1 = 6 and a + 3d = 5 + 3 = 8 Ans; 2, 4, 6 and 8. check: 2 + 4 + 6 + 8 = 20 (2x 8)/(4 x 6) = 16/24 = 2/3 read less
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2 , 4 , 6 , 8 2 x 8 = 16 and 4 x 6 = 24 so ratio 2 : 3
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let us take 4 nos as x-(3a/2), x-(a/2), x+(3a/2), x+(a/2). adding these we get 4x=20, or, x=5. then acc to given condition, we get a=2. then putting all values we get all he 4 nos.
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let the four numbers to be the a-3d ,a-d, a+d ,a+3d as the numbers are in arithmetic progressions with common difference: 2d and (a-3d) +(a-d)+(a+d) +(a+3d)=20 we get 4a=20 a=5 (a-3d)(a+3d)/(a+d)(a-d)=2/3 (a-3d)(a+3d)*3=(a+d)(a-d)*2 i.e. 3a^2-27d^2=2a^2-2d^2 a^2=25d^2 :now substitute a=5 we...
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let the four numbers to be the a-3d ,a-d, a+d ,a+3d as the numbers are in arithmetic progressions with common difference: 2d and (a-3d) +(a-d)+(a+d) +(a+3d)=20 we get 4a=20 a=5 (a-3d)(a+3d)/(a+d)(a-d)=2/3 (a-3d)(a+3d)*3=(a+d)(a-d)*2 i.e. 3a^2-27d^2=2a^2-2d^2 a^2=25d^2 :now substitute a=5 we get 25=25d^2 d^2=1 d=+1,-1 so the numbers become 2,4,6,8 read less
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let the numbers be a-3d, a-d, a+d, a+3d, given their sum is 20. a-3d+a-d+a+d+a+3d=20 after solving, a=5....(1) we know (a-3d)(a-3d)/(a-d)(a+d) = 2/3...(2) from (1), substitute the value of a in (2) by solving we will get d=+1 or -1 after substituting the values of a & d, we get the numbers...
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let the numbers be a-3d, a-d, a+d, a+3d, given their sum is 20. a-3d+a-d+a+d+a+3d=20 after solving, a=5....(1) we know (a-3d)(a-3d)/(a-d)(a+d) = 2/3...(2) from (1), substitute the value of a in (2) by solving we will get d=+1 or -1 after substituting the values of a & d, we get the numbers as 2,4,6,8 read less
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let the numbers be a-3d,a-d,a+d,a+3d. a-3d+a-d+a+d+a+3d=20 a=5 (5-3d)(5+3d)/(5-d)(5+d)=2/3 25-9d2/25-d2 =2/3 75-27d2=50-2d2 25d2=25 d=1 or d=-1.....now proceed
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