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The sum of three numbers in a GP is 14 and the sum of their squares is 84. Find the largestnumber.
- a)8
- b)6
- c)4
- d)12
Correct answer is option 'A'. Can you explain this answer?
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The sum of three numbers in a GP is 14 and the sum of their squares is...
Visualising the squares below 84, we can see that the only way to get the sum of 3 squares as 84
is: 22 + 42 + 82 = 4 + 16 + 64. The largest number is 8. Option (a) is correct.
is: 22 + 42 + 82 = 4 + 16 + 64. The largest number is 8. Option (a) is correct.
Most Upvoted Answer
The sum of three numbers in a GP is 14 and the sum of their squares is...
To solve this problem, we can use the formulas for the sum of a geometric progression (GP) and the sum of the squares of a GP.
Let's assume that the three numbers in the GP are "a/r", "a", and "ar", where "a" is the first term and "r" is the common ratio.
The sum of the three numbers is given as 14:
a/r + a + ar = 14 ----(1)
The sum of their squares is given as 84:
(a/r)^2 + a^2 + (ar)^2 = 84 ----(2)
Now, let's simplify equations (1) and (2) to solve for "a" and "r".
Simplifying equation (1):
a/r + a + ar = 14
a(1/r + 1 + r) = 14
a(r^2 + r + 1)/r = 14
a(r^2 + r + 1) = 14r ----(3)
Simplifying equation (2):
(a/r)^2 + a^2 + (ar)^2 = 84
a^2/r^2 + a^2 + a^2r^2 = 84
a^2(1/r^2 + 1 + r^2) = 84
a^2(r^2 + 1 + r^4)/r^2 = 84
a^2(r^4 + r^2 + 1) = 84r^2 ----(4)
Now, let's substitute the value of "a" from equation (3) into equation (4).
(a(r^2 + r + 1))^2 = 84r^2
(a^2(r^2 + r + 1)^2) = 84r^2
[(14r)/(r^2 + r + 1)]^2(r^2 + r + 1)^2 = 84r^2
(14r)^2 = 84r^2
196r^2 = 84r^2
112r^2 = 0
r^2 = 0
Since the common ratio "r" cannot be zero, this means that there is no valid solution for this problem. Therefore, the given information is inconsistent and there is no largest number that satisfies both conditions.
Hence, the correct answer would be considered as "None of the above" or "Not possible to determine".
Let's assume that the three numbers in the GP are "a/r", "a", and "ar", where "a" is the first term and "r" is the common ratio.
The sum of the three numbers is given as 14:
a/r + a + ar = 14 ----(1)
The sum of their squares is given as 84:
(a/r)^2 + a^2 + (ar)^2 = 84 ----(2)
Now, let's simplify equations (1) and (2) to solve for "a" and "r".
Simplifying equation (1):
a/r + a + ar = 14
a(1/r + 1 + r) = 14
a(r^2 + r + 1)/r = 14
a(r^2 + r + 1) = 14r ----(3)
Simplifying equation (2):
(a/r)^2 + a^2 + (ar)^2 = 84
a^2/r^2 + a^2 + a^2r^2 = 84
a^2(1/r^2 + 1 + r^2) = 84
a^2(r^2 + 1 + r^4)/r^2 = 84
a^2(r^4 + r^2 + 1) = 84r^2 ----(4)
Now, let's substitute the value of "a" from equation (3) into equation (4).
(a(r^2 + r + 1))^2 = 84r^2
(a^2(r^2 + r + 1)^2) = 84r^2
[(14r)/(r^2 + r + 1)]^2(r^2 + r + 1)^2 = 84r^2
(14r)^2 = 84r^2
196r^2 = 84r^2
112r^2 = 0
r^2 = 0
Since the common ratio "r" cannot be zero, this means that there is no valid solution for this problem. Therefore, the given information is inconsistent and there is no largest number that satisfies both conditions.
Hence, the correct answer would be considered as "None of the above" or "Not possible to determine".
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The sum of three numbers in a GP is 14 and the sum of their squares is 84. Find the largestnumber.a)8b)6c)4d)12Correct answer is option 'A'. Can you explain this answer?
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