Question

Divide 56 in four parts in A.P. Such that the ratio of the product of their extremes ($1^{st}$ and $4^{th}$) to the product of means ($2^{nd}$ and $3^{rd}$) is 5:6.

Answer 1

Hint: In this question it is given that, divide 56 in four parts in A.P. Such that the ratio of the product of their extremes ($1^{st}$ and $4^{th}$) to the product of means ($2^{nd}$ and $3^{rd}$) is 5:6. So for finding a solution we have to take the 4 parts in such a way that they are in A.P, i.e, the difference of consecutive two terms are equal.

Complete step-by-step solution:
Let us consider the 4 parts be (a-3d), (a-d), (a+d), (a+3d), where we take common difference as 2d,
i.e, (${2}^{nd}$ term - ${1}^{st}$ term)=(a-d)-(a-3d)=2d.
Now, since the sum of these four parts is 56,
Therefore,$\left( a-3d\right) +\left( a-d\right) +\left( a+d\right) +\left( a+3d\right) =56$
$\Rightarrow 4a=56$
$\Rightarrow a=\dfrac{56}{4}$
$\Rightarrow a=14$

Now also it is given that, the ratio of the product of their extremes ($1^{st}$ and $4^{th}$) to the product of means ($2^{nd}$ and $3^{rd}$) is 5:6.
So we can write,
$\dfrac{\left( a-3d\right) \left( a+3d\right) }{\left( a-d\right) \left( a+d\right) } =\dfrac{5}{6}$
$\Rightarrow \dfrac{a^{2}-\left( 3d\right)^{2} }{a^{2}-d^{2}} =\dfrac{5}{6}$
$\Rightarrow \dfrac{a^{2}-9d^{2}}{a^{2}-d^{2}} =\dfrac{5}{6}$
$\Rightarrow 6a^{2}-54d^{2}=5a^{2}-5d^{2}$
$\Rightarrow 6a^{2}-5a^{2}=54d^{2}-5d^{2}$
$\Rightarrow a^{2}=49d^{2}$
$\Rightarrow a=\pm \sqrt{49d^{2}}$
$\Rightarrow a=\pm \sqrt{\left( 7d\right)^{2} }$
$\Rightarrow a=\pm 7d$
$\Rightarrow 14=\pm 7d$ [since a=14]
$\Rightarrow d=\pm 2$
Now we take d=2,
So the four terms of A.P will be,

\Rightarrow \left( 14-3\times 2\right) ,\left( 14-2\right) ,\left( 14+2\right) ,\left( 14+3\times 2\right) $$ $$\therefore 8,12,16,20$$ Note: To solve this type of question you need to have the basic idea about A.P and how to take terms which are in A.P. So apart from (a-3d), (a-d), (a+d), (a+3d), you can also take different terms which are in A.P. And in the last part of the solution we take d=2, but if you take d=-2 then this will also give you the same terms.