 QUESTION

# A number 32 which is divided into four parts which are in A.P. such that the ratio of product of extremes to the product of means is 7:15. Then find the four parts.

Hint- In this question, we use the concept of Arithmetic Progression. As given in the question, we’ll assume 4 parts and add them to equate it to 32. Then we’ll find the first number of the series. With that we can proceed to find the value of ‘d’ by using the ratio of the product of extremes to the product of means and reach the final answer.

Let the four parts be $\left( {a - 3d} \right),\left( {a - d} \right),\left( {a + d} \right){\text{ and }}\left( {a + 3d} \right)$ which are in A.P.
Given that the sum of all parts of a number is 32.
$\Rightarrow \left( {a - 3d} \right) + \left( {a - d} \right) + \left( {a + d} \right) + \left( {a + 3d} \right) = 32 \\ \Rightarrow 4a = 32 \\ \Rightarrow a = \dfrac{{32}}{4} \\ \Rightarrow a = 8 \\$
Now, we have to find the value of d by using the ratio of the product of extremes to the product of means is 7:15.
Product of extreme $= \left( {a - 3d} \right)\left( {a + 3d} \right)$
Product of means $= \left( {a - d} \right)\left( {a + d} \right)$
Now, the ratio of the product of extremes to the product of means is 7:15.
$\Rightarrow \dfrac{{\left( {a - 3d} \right)\left( {a + 3d} \right)}}{{\left( {a - d} \right)\left( {a + d} \right)}} = \dfrac{7}{{15}}$
Using the identity $\left( {x + y} \right)\left( {x - y} \right) = {x^2} - {y^2}$
$\Rightarrow \dfrac{{{a^2} - 9{d^2}}}{{{a^2} - {d^2}}} = \dfrac{7}{{15}}$
Apply cross multiplication,
$\Rightarrow 15\left( {{a^2} - 9{d^2}} \right) = 7\left( {{a^2} - {d^2}} \right) \\ \Rightarrow 15{a^2} - 135{d^2} = 7{a^2} - 7{d^2} \\ \Rightarrow 135{d^2} - 7{d^2} = 15{a^2} - 7{a^2} \\ \Rightarrow 128{d^2} = 8{a^2} \\$
Now put the value of a=8
$\Rightarrow {d^2} = \dfrac{{8 \times {{\left( 8 \right)}^2}}}{{128}} \\ \Rightarrow {d^2} = \dfrac{{512}}{{128}} \\ \Rightarrow {d^2} = 4 \\$
Taking square root,
$\Rightarrow d = \pm 2$

Case 1: a=8 and d=2
$\left( {a - 3d} \right) = 8 - 3 \times 2 = 2 \\ \left( {a - d} \right) = 8 - 2 = 6 \\ \left( {a + d} \right) = 8 + 2 = 10 \\ \left( {a + 3d} \right) = 8 + 3 \times 2 = 14 \\$
Four parts are 2, 6, 10 and 14.

Case 2: a=8 and d=-2
$\left( {a - 3d} \right) = 8 - 3 \times \left( { - 2} \right) = 14 \\ \left( {a - d} \right) = 8 - \left( { - 2} \right) = 10 \\ \left( {a + d} \right) = 8 + \left( { - 2} \right) = 6 \\ \left( {a + 3d} \right) = 8 + 3 \times \left( { - 2} \right) = 2 \\$
Four parts are 14, 10, 6 and 2.

So, the four parts are 2, 6 , 10 and 14.

Note-In such types of problems we use some important points to solve questions in an easy way. First we assume four parts which are in A.P. and we know the sum of all parts is equal to 32. So, we get the value of a. Then we use the ratio mentioned in the question, we know the product of extremes is (a-3d)(a+3d) and the product of means is (a-d)(a+d). After using ratio we will get the value of d. Then putting the value of a and d in four parts so we will get the required four parts.