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The sum of three numbers in G.P. is 70. If the two extremes by multiplied each by 4 and the mean by 5, the products are in AP. The numbers are
- a)12, 18, 40
- b)10, 20, 40
- c)40, 20, 10
- d)none of these
Correct answer is option 'B,C'. Can you explain this answer?
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The sum of three numbers in G.P. is 70. If the two extremes by multipl...
Given information:
- Sum of three numbers in G.P. is 70.
- If the two extremes by multiplied each by 4 and the mean by 5, the products are in AP.
Let the three numbers be a, ar and ar^2, where r is the common ratio.
Finding the values of a and r:
a + ar + ar^2 = 70 (sum of three numbers is 70)
a(1 + r + r^2) = 70
a = 70 / (1 + r + r^2)
Using the second condition:
4a * 4ar * 5(ar) is in AP
20ar^3, 80ar^2, 80a are in AP
80ar^2 - 20ar^3 = 80a - 80ar^2
4r(5 - r)(a) = 4(20 - 4r)(a)
r(5 - r) = 5 - r
r^2 - 6r + 5 = 0
(r - 1)(r - 5) = 0
r = 1 or r = 5
If r = 1, then a = 70 / 3, ar = 70 / 3 and ar^2 = 70 / 3
But this is not a valid G.P. as all the terms are equal.
If r = 5, then a = 2, ar = 10 and ar^2 = 50
These are the required numbers in G.P.
Checking the second condition:
4a * 4ar * 5(ar) is in AP
(32, 160, 200) is in AP
160 - 32 = 200 - 160
128 = 40
Therefore, the answer is option B and C (10, 20, 40 and 40, 20, 10).
- Sum of three numbers in G.P. is 70.
- If the two extremes by multiplied each by 4 and the mean by 5, the products are in AP.
Let the three numbers be a, ar and ar^2, where r is the common ratio.
Finding the values of a and r:
a + ar + ar^2 = 70 (sum of three numbers is 70)
a(1 + r + r^2) = 70
a = 70 / (1 + r + r^2)
Using the second condition:
4a * 4ar * 5(ar) is in AP
20ar^3, 80ar^2, 80a are in AP
80ar^2 - 20ar^3 = 80a - 80ar^2
4r(5 - r)(a) = 4(20 - 4r)(a)
r(5 - r) = 5 - r
r^2 - 6r + 5 = 0
(r - 1)(r - 5) = 0
r = 1 or r = 5
If r = 1, then a = 70 / 3, ar = 70 / 3 and ar^2 = 70 / 3
But this is not a valid G.P. as all the terms are equal.
If r = 5, then a = 2, ar = 10 and ar^2 = 50
These are the required numbers in G.P.
Checking the second condition:
4a * 4ar * 5(ar) is in AP
(32, 160, 200) is in AP
160 - 32 = 200 - 160
128 = 40
Therefore, the answer is option B and C (10, 20, 40 and 40, 20, 10).
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The sum of three numbers in G.P. is 70. If the two extremes by multiplied each by 4 and the mean by 5, the products are in AP. The numbers area)12, 18, 40b)10, 20, 40c)40, 20, 10d)none of theseCorrect answer is option 'B,C'. Can you explain this answer?
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