# 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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Updated on: 10-Oct-2022

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