Four numbers are in G.P. If sum of first two numbers is 44 and sum of the last two numbers is 396 then ,find the numbers
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Let the 4 numbers are a/r^3,a/r, ar, ar^3
According to question
a/r^3+a/r=44
Or, (a/r^3)×(1+r^2)=44....(1)
Again ar+ar^3=396
Or, ar(1+r^2)=396...(2)
dividing (2) by (1) we have
ar(1+r^2)/{a(1+r^2)/r^3}=396/44
Or, r^4==9
Or, r=+-√3
taking r=+√3 we have from(1)
(a/3√3)×(1+3)=44
Or, a=(44×3√3)/4
Or a=33√3
Hence the numbers are
33√3/3√3, 33√3/√3, 33√3×√3, 33√3×3√3
i.e 11,33,99, 297
taking r=-√3 we get another set of numbers.
According to question
a/r^3+a/r=44
Or, (a/r^3)×(1+r^2)=44....(1)
Again ar+ar^3=396
Or, ar(1+r^2)=396...(2)
dividing (2) by (1) we have
ar(1+r^2)/{a(1+r^2)/r^3}=396/44
Or, r^4==9
Or, r=+-√3
taking r=+√3 we have from(1)
(a/3√3)×(1+3)=44
Or, a=(44×3√3)/4
Or a=33√3
Hence the numbers are
33√3/3√3, 33√3/√3, 33√3×√3, 33√3×3√3
i.e 11,33,99, 297
taking r=-√3 we get another set of numbers.
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