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divide 56 in four parts in A.P. such that the ratio of the product of their extremes to the product of their means is 5 : 6.
Given:
The ratio of the product of the extremes to the product of the means is 5 : 6.
To do:
We have to divide 56 in four parts in A.P. such that the ratio of the product of their extremes to the product of their means is 5 : 6.
Solution:
Let the four numbers of the A.P. be $(a-3 d), (a-d), (a+d), (a+3 d)$
Here,
Extremes are $(a-3 d)$ and $(a+3 d)$
Means are $(a-d)$ and $(a+d)$
Product of extremes $=(a-3 d)\times(a+3 d)$
$=(a)^2-(3d)^2$ (Since $(a+b)(a-b)=a^2-b^2$)
$=a^2-9d^2$
Product of mean $=(a- d)\times(a+ d)$
$=(a)^2-(d)^2$ (Since $(a+b)(a-b)=a^2-b^2$)
$=a^2-d^2$
According to the question,
The ratio of the product of their extremes to the product of their means is 5 : 6.
This implies,
$a^2-9d^2:a^2-9d^2=5:6$
$\Rightarrow \frac{a^2-9d^2}{a^2-d^2}=\frac{5}{6}$
On cross multiplication, we get,
$6(a^{2}-9 d^{2})=5(a^{2}-d^{2})$
$6(a^2)-6(9d^2)=5(a^2)-5(d^2)$
$6 a^{2}-54 d^{2}=5 a^{2}-5 d^{2}$
$6 a^{2}-5 a^{2}=54 d^{2}-5 d^{2}$
$a^{2}=49 d^{2}$
$(a)^{2}=(7 d)^{2}$ (Since $49=7^2$)
$a=7 d$
Therefore,
Given that,
$(a-3 d)+(a-d)+(a+d)+(a+3 d)=56$
$a+a+a+a-3d-d+d+3d=56$
$4a=56$
$a=\frac{56}{4}$
$a=14$
Substituting $a=7d$ in the above equation, we get,
$14=7d$
$d=\frac{14}{7}$
$d=2$
The required numbers are,
$a-3d=14-3(2)=14-6=8$
$a-d=14-2=12$
$a+d=14+2=16$
$a+3d=14+3(2)=14+6=20$
The required four numbers in A.P. are $8, 12, 16$ and $20$.
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