Question

If average of four numbers is 20 and average of their squares is 500, then average of their products taken two at a time is​

Answer 1

Answer:

Average of their product taken two at a time

is 366.67

Step-by-step explanation:

Formula used:

$(a+b+c+d)^2=a^2+b^2+c^2+d^2+2(ab+bc+cd+ad+bd+ac)$

Let the fours be a, b, c, d

Then,

$\frac{a+b+c+d}{4}=20$

$a+b+c+d=80$

Also,

$\frac{a^2+b^2+c^2+d^2}{4}=500$

$a^2+b^2+c^2+d^2=2000$

Now,

$(a+b+c+d)^2=a^2+b^2+c^2+d^2+2(ab+bc+cd+ad+bd+ac)$

$(80)^2=2000+2(ab+bc+cd+ad+bd+ac)$

$6400=2000+2(ab+bc+cd+ad+bd+ac)$

$6400-2000=2(ab+bc+cd+ad+bd+ac)$

$4400=2(ab+bc+cd+ad+bd+ac)$

$2200=ab+bc+cd+ad+bd+ac$

Average of their product taken two at a time

$=\frac{ab+bc+cd+ad+bd+ac}{6}$

$=\frac{2200}{6}$

$=\frac{1100}{3}$

$=366.67$

Answer 2

Answer:

Average of their products taken two at a time is​ 366.67.

Step-by-step explanation:

Given : If average of four numbers is 20 and average of their squares is 500.

To find : Average of their products taken two at a time is ?

Solution :

Let the four numbers be a, b, c, d

According to question,

$\frac{a+b+c+d}{4}=20\\\\a+b+c+d=80$....(1)

$\frac{a^2+b^2+c^2+d^2}{4}=500\\\\a^2+b^2+c^2+d^2=2000$.....(2)

We know the identity,

$(a+b+c+d)^2=a^2+b^2+c^2+d^2+2(ab+bc+cd+ad+bd+ac)$

Substituting the values we get,

$(80)^2=2000+2(ab+bc+cd+ad+bd+ac)\\6400=2000+2(ab+bc+cd+ad+bd+ac)\\6400-2000=2(ab+bc+cd+ad+bd+ac)\\4400=2(ab+bc+cd+ad+bd+ac)\\2200=ab+bc+cd+ad+bd+ac$

Now, we find the average of their products taken two time is

$=\frac{ab+bc+cd+ad+bd+ac}{6}\\\\=\frac{2200}{6}\\\\=\frac{1100}{3}\\\\=366.67$

Therefore, Average of their products taken two at a time is​ 366.67.