Divide 72 in four parts in A.P, such that the ratio of the product of their extremes (1st and 4th ) to the product of means (2nd and 3rd) is 27/35 Find the four parts.
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Divide 72 in four parts in A.P, such that the ratio of the product of their extremes (1st and 4th ) to the product of means (2nd and 3rd) is 27/35 Find the four parts.
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Answer:
Let the four parts be (a−3d),(a−d),(a+d) and (a+3d).
Then, Sum of the numbers =32
⟹(a−3d)+(a−d)+(a+d)+(a+3d)=32⟹4a=32⟹a=8
It is given that
(a−d)(a+d)
(a−3d)(a+3d)
=
15
7
⟹
a
2
−d
2
a
2
−9d
2
=
15
7
⟹
64−d
2
64−9d
2
=
15
7
⟹128d
2
=512⟹d
2
=4⟹d=±2
Thus, the four parts are a−3d,a−d,a+d and 3d, i.e. 2,6,10,14.
Step-by-step explanation:
Answer:
The four parts given AP are 9,15,21,27
Step-by-step explanation:
Step 1 of 1:
[tex]a-3d,a-d,a+d,a+3d=72\\4a=72\\a=18[/tex]
[tex]\frac{(a-3d)(a+3d)}{(a-d)(a+d)} =\frac{27}{35} \\\frac{a^2-9d^2}{a^2-d^2} =\frac{27}{35} \\\\35a^2-(35*9)d^2=27a^2-27d^2\\35a^2-315d^2=27a^2-27d^2\\\\8a^2=288d^2\\288d^2=8*18^2\\288d^2=2592\\d^2=9\\d=-3,3[/tex]