Divide 32 into
4 pairs which are the
4 term of an Ap such
that the product
of the
1st and the
4 th term is to the product of the 2nd and the 3rd terms
as 7:15
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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:
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Answer:
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