Find the 3 numbers in GP is whose sum is 13and the sum of whose squares is 91
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Let the three numbers be a, ar, ar^2.
Given that the sum of three numbers is 13.
= > a + ar + ar^2 = 13 --------- (1)
Given that sum of their squares is 91.
= > a^2 + a^2r^2 + a^2r^4 = 91 ------- (2)
On Squaring equation (1) on both sides, we get
= > (a + ar + ar^2)^2 = (13)^2
We know that (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca
= > a + a^2r^2 + a^2r^4 + 2a^2r + 2a^2r^3 + 2a^2r^2 = 169
= > 91 + 2a^2r + 2a^2r^3 + 2a^2r^2 = 169
= > 2ar(a + ar + ar^2) = 169
= > 2ar(13) = 169 - 91
= > 26ar = 78
= > ar = 3.
The values which satisfy the equations are a = 1, r = 3 (or) 3,1
Now,
a = 1 (or) 3
ar = 3
ar^2 = 9 (or) 3
Therefore the three numbers are (1,3,9) (or) (9,3,1)
Hope this helps!
Given that the sum of three numbers is 13.
= > a + ar + ar^2 = 13 --------- (1)
Given that sum of their squares is 91.
= > a^2 + a^2r^2 + a^2r^4 = 91 ------- (2)
On Squaring equation (1) on both sides, we get
= > (a + ar + ar^2)^2 = (13)^2
We know that (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca
= > a + a^2r^2 + a^2r^4 + 2a^2r + 2a^2r^3 + 2a^2r^2 = 169
= > 91 + 2a^2r + 2a^2r^3 + 2a^2r^2 = 169
= > 2ar(a + ar + ar^2) = 169
= > 2ar(13) = 169 - 91
= > 26ar = 78
= > ar = 3.
The values which satisfy the equations are a = 1, r = 3 (or) 3,1
Now,
a = 1 (or) 3
ar = 3
ar^2 = 9 (or) 3
Therefore the three numbers are (1,3,9) (or) (9,3,1)
Hope this helps!
siddhartharao77:
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