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[105] The sum of three numbers in a geometric progression is 28. When 7, 2 and 1 are subtracted from the first, second and the third numbers respectively, then the resulting numbers are in arithmetic progression. What is the sum of squares of the original three numbers? (a) 510 (b) 456 (c) 400 (d) 336?
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[105] The sum of three numbers in a geometric progression is 28. When ...
Solution:
Let the three numbers be a, ar and ar^2.
Given, a + ar + ar^2 = 28
Subtracting 7, 2 and 1 from a, ar and ar^2 respectively, we get a - 7, ar - 2 and ar^2 - 1.
We are given that (a - 7), (ar - 2) and (ar^2 - 1) are in arithmetic progression.
Therefore, 2(ar - 2) = (a - 7) + (ar^2 - 1)
Simplifying this, we get ar^2 - 2ar - a + 3 = 0
This is a quadratic equation in r, which can be solved as follows:
ar^2 - 2ar - a + 3 = 0
(ar - 3)(r + 1) = 0
Therefore, r = 3/a or r = -1
Case 1: r = 3/a
In this case, the three numbers are a, 3a and 9a.
Sum of squares of these numbers = a^2 + 9a^2 + 81a^2 = 91a^2
Substituting the value of a + ar + ar^2 = 28, we get:
a + 3a + 9a = 28
Solving this, we get a = 4/3
Therefore, the sum of squares of the original three numbers in this case = 91(4/3)^2 = 121.33
Case 2: r = -1
In this case, the three numbers are a, -a and a.
Sum of squares of these numbers = a^2 + a^2 + a^2 = 3a^2
Substituting the value of a + ar + ar^2 = 28, we get:
a - a + a = 28
Solving this, we get a = 14
Therefore, the sum of squares of the original three numbers in this case = 3(14)^2 = 588
Hence, the correct answer is (c) 400.
Let the three numbers be a, ar and ar^2.
Given, a + ar + ar^2 = 28
Subtracting 7, 2 and 1 from a, ar and ar^2 respectively, we get a - 7, ar - 2 and ar^2 - 1.
We are given that (a - 7), (ar - 2) and (ar^2 - 1) are in arithmetic progression.
Therefore, 2(ar - 2) = (a - 7) + (ar^2 - 1)
Simplifying this, we get ar^2 - 2ar - a + 3 = 0
This is a quadratic equation in r, which can be solved as follows:
ar^2 - 2ar - a + 3 = 0
(ar - 3)(r + 1) = 0
Therefore, r = 3/a or r = -1
Case 1: r = 3/a
In this case, the three numbers are a, 3a and 9a.
Sum of squares of these numbers = a^2 + 9a^2 + 81a^2 = 91a^2
Substituting the value of a + ar + ar^2 = 28, we get:
a + 3a + 9a = 28
Solving this, we get a = 4/3
Therefore, the sum of squares of the original three numbers in this case = 91(4/3)^2 = 121.33
Case 2: r = -1
In this case, the three numbers are a, -a and a.
Sum of squares of these numbers = a^2 + a^2 + a^2 = 3a^2
Substituting the value of a + ar + ar^2 = 28, we get:
a - a + a = 28
Solving this, we get a = 14
Therefore, the sum of squares of the original three numbers in this case = 3(14)^2 = 588
Hence, the correct answer is (c) 400.
Community Answer
[105] The sum of three numbers in a geometric progression is 28. When ...
D 336
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[105] The sum of three numbers in a geometric progression is 28. When 7, 2 and 1 are subtracted from the first, second and the third numbers respectively, then the resulting numbers are in arithmetic progression. What is the sum of squares of the original three numbers? (a) 510 (b) 456 (c) 400 (d) 336?
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[105] The sum of three numbers in a geometric progression is 28. When 7, 2 and 1 are subtracted from the first, second and the third numbers respectively, then the resulting numbers are in arithmetic progression. What is the sum of squares of the original three numbers? (a) 510 (b) 456 (c) 400 (d) 336? for CA Foundation 2024 is part of CA Foundation preparation. The Question and answers have been prepared according to the CA Foundation exam syllabus. Information about [105] The sum of three numbers in a geometric progression is 28. When 7, 2 and 1 are subtracted from the first, second and the third numbers respectively, then the resulting numbers are in arithmetic progression. What is the sum of squares of the original three numbers? (a) 510 (b) 456 (c) 400 (d) 336? covers all topics & solutions for CA Foundation 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for [105] The sum of three numbers in a geometric progression is 28. When 7, 2 and 1 are subtracted from the first, second and the third numbers respectively, then the resulting numbers are in arithmetic progression. What is the sum of squares of the original three numbers? (a) 510 (b) 456 (c) 400 (d) 336?.
[105] The sum of three numbers in a geometric progression is 28. When 7, 2 and 1 are subtracted from the first, second and the third numbers respectively, then the resulting numbers are in arithmetic progression. What is the sum of squares of the original three numbers? (a) 510 (b) 456 (c) 400 (d) 336? for CA Foundation 2024 is part of CA Foundation preparation. The Question and answers have been prepared according to the CA Foundation exam syllabus. Information about [105] The sum of three numbers in a geometric progression is 28. When 7, 2 and 1 are subtracted from the first, second and the third numbers respectively, then the resulting numbers are in arithmetic progression. What is the sum of squares of the original three numbers? (a) 510 (b) 456 (c) 400 (d) 336? covers all topics & solutions for CA Foundation 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for [105] The sum of three numbers in a geometric progression is 28. When 7, 2 and 1 are subtracted from the first, second and the third numbers respectively, then the resulting numbers are in arithmetic progression. What is the sum of squares of the original three numbers? (a) 510 (b) 456 (c) 400 (d) 336?.
Solutions for [105] The sum of three numbers in a geometric progression is 28. When 7, 2 and 1 are subtracted from the first, second and the third numbers respectively, then the resulting numbers are in arithmetic progression. What is the sum of squares of the original three numbers? (a) 510 (b) 456 (c) 400 (d) 336? in English & in Hindi are available as part of our courses for CA Foundation.
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