Question

Find four proportions such that the sum of the extremes is $21$.The sum of the means $19$, and the sum of the squares of all the four numbers is $442$.

Answer 1

Hint : In this question we will use the concept of ratio and proportion. Here we have given some statements in the question ,we will use those statements and make some equations to solve this problem . by using those equations and some properties of ratio and proportion we will get our answer.

Complete step-by-step answer: -
Given that , sum of extremes =$21$, sum of means =$19$ and the sum of the squares of all four numbers =$442$
Let us assume the four numbers are $a,b,c{\text{ and }}d$.
According to the question , sum of extremes =$21$
So,
$ \Rightarrow a + d = 21$ ………(i)
Sum of means =$19$, so
$ \Rightarrow b + c = 19$ ……….(ii)
Sum of squares of all four numbers =$442$, so
$ \Rightarrow ({a^2} + {b^2} + {c^2} + {d^2}) = 442$ .
As given, these numbers are in proportion, so
$ad = bc$.
Now we know that , when the ratio of 2 terms = ratio of the other two terms ,then they are in proportion .
So, by squaring and adding equations (i) and (ii) , we get
$
   \Rightarrow {(a + d)^2} + {(b + c)^2} = {(21)^2} + {(19)^2}. \\
   \Rightarrow \left( {{a^2} + {d^2} + 2ad} \right) + ({b^2} + {c^2} + 2bc) = 441 + 361 \\
   \Rightarrow ({a^2} + {b^2} + {c^2} + {d^2} + 2ad + 2bc) = 802 \\
$
We have , $ad = bc{\text{ and }}({a^2} + {b^2} + {c^2} + {d^2}) = 442$ ,then
$
   \Rightarrow 442 + 4ad = 802 \\
   \Rightarrow 4ad = 802 - 442 \\
   \Rightarrow 4ad = 360 \\
   \Rightarrow ad = \dfrac{{360}}{4} \\
   \Rightarrow ad = 90 \\
$
We know that , $ad = bc{\text{ }}$
So, $bc = 90$.
Let us take $[{(a - d)^2} = {(a + d)^2} - 4ad]$, then
$
   \Rightarrow {(a - d)^2} = {(21)^2} - 4 \times 90 \\
   \Rightarrow {(a - d)^2} = 441 - 360 \\
   \Rightarrow {(a - d)^2} = 81 \\
   \Rightarrow (a - d) = \sqrt {81} \\
$
$ \Rightarrow (a - d) = 9$ ………….(iii)
Now, adding equations (i) and (iii), we get
$
   \Rightarrow (a + d) + (a - d) = 21 + 9 \\
   \Rightarrow 2a = 30 \\
   \Rightarrow a = \dfrac{{30}}{2} \\
   \Rightarrow a = 15 \\
$
Put this value of a in equation (i),
$
   \Rightarrow (a + d) = 21 \\
   \Rightarrow d = 21 - a \\
   \Rightarrow d = 21 - 15 \\
   \Rightarrow d = 6. \\
$
Now similarly we will take , $[{(b - c)^2} = {(b + c)^2} - 4bc]$, then
$
   \Rightarrow {(b - c)^2} = {(19)^2} - 4 \times 90 \\
   \Rightarrow {(b - c)^2} = 361 - 360 \\
   \Rightarrow {(b - c)^2} = 1 \\
   \Rightarrow (b - c) = \sqrt 1 \\
$
$ \Rightarrow (b - c) = 1$ ……….(iv)
Now, adding equations (ii) and (iv), we get
$
   \Rightarrow (b + c) + (b - c) = 19 + 1 \\
   \Rightarrow 2b = 20 \\
   \Rightarrow b = \dfrac{{20}}{2} \\
   \Rightarrow b = 10 \\
$
Now ,put this value in equation (ii), we get
$
   \Rightarrow (b + c) = 19 \\
   \Rightarrow c = 21 - b \\
   \Rightarrow c = 19 - 10 \\
   \Rightarrow c = 9 \\
$
Hence, the four numbers are $15,10,9{\text{ and }}6$.

Note : In this type of question we will use concepts and properties of ratio and proportion. First ,we will make some equations from the statements given in the question then by using the properties in those equations we will solve them one by one and after solving them we will get all the required numbers .