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The product of 3 numbers in G P is 729 and the sum of squares is 819. The numbers are
- a)9, 3, 27
- b)27, 3, 9
- c)3, 9, 27
- d)none of these
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
The product of 3 numbers in G P is 729 and the sum of squares is 819. ...
Given:
Product of three numbers in G.P = 729
Sum of squares of three numbers = 819
To find:
The three numbers
Solution:
Let the three numbers in G.P be a, ar, and ar^2
Given, a * ar * ar^2 = 729
=> a^3 r^3 = 729
=> ar = 9
Substituting ar = 9, we get a and ar^2 as 3 and 27 respectively.
Now, we need to check if the sum of squares of these three numbers is equal to 819.
Sum of squares = a^2 + (ar)^2 + (ar^2)^2
= a^2 + a^2 r^2 + a^2 r^4
= a^2 (1 + r^2 + r^4)
Substituting a = 3 and r = 3, we get
Sum of squares = 3^2 (1 + 3^2 + 3^4) = 9 * 91 = 819
Therefore, the three numbers in G.P are 3, 9, and 27, and the correct option is (c).
Product of three numbers in G.P = 729
Sum of squares of three numbers = 819
To find:
The three numbers
Solution:
Let the three numbers in G.P be a, ar, and ar^2
Given, a * ar * ar^2 = 729
=> a^3 r^3 = 729
=> ar = 9
Substituting ar = 9, we get a and ar^2 as 3 and 27 respectively.
Now, we need to check if the sum of squares of these three numbers is equal to 819.
Sum of squares = a^2 + (ar)^2 + (ar^2)^2
= a^2 + a^2 r^2 + a^2 r^4
= a^2 (1 + r^2 + r^4)
Substituting a = 3 and r = 3, we get
Sum of squares = 3^2 (1 + 3^2 + 3^4) = 9 * 91 = 819
Therefore, the three numbers in G.P are 3, 9, and 27, and the correct option is (c).
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