Question

The average of the squares of five consecutive positive integers is 66. Find the average of these five integers.​

Answer 1

$Let \: x , (x+1),(x+2),(x+3)\: and \:(x+4)\:are \\five\: consecutive \: positive \: integers .$

$Average \: of \: the \: squares \: of \: these \\five \: consecutive \: integers = 66 \:(given)$

$\implies \frac{x^{2}+(x+1)^{2}+(x+2)^{2}+(x+3)^{2}+(x+4)^{2}}{5} = 66$

$\implies \frac{5x^{2} + 20x + 30 }{5} - 66 = 0$

$\implies \frac{5(x^{2}+4x+6}{5} - 66 = 0$

$\implies x^{2} + 4x + 6 - 66 = 0$

$\implies x^{2} + 4x - 60 = 0$

/* Splitting the middle term,we get */

$\implies x^{2} + 10x - 6x - 60 = 0$

$\implies x( x + 10 ) - 6( x + 10 ) = 0$

$\implies ( x + 10 )( x - 6 ) = 0$

$\implies x + 10 = 0 \: Or \: x - 6 = 0$

$\implies x = - 10 \: Or \: x = 6$

/* It is given that x is positive integer */

$x = 6$

$Now, Five \: consecutive \: numbers \:are \\6, (6+1),(6+2),(6+3)\:and \:(6+4) \\i.e ., 6,7,8,9\:and \:10$

$Average \: of \: 5 \: integers \\= \frac{6+7+8+9+10}{5}\\= \frac{40}{5}\\= 8$

Therefore.,

$\red {Average \: of \: 5 \: integers}\green {= 8}$

•••♪

Answer 2

$\huge\bold\green{Question}$

The average of the squares of five consecutive positive integers is 66. Find the average of these five integers.

$\huge\bold\green{Answer}$

Let us assumed five consecutive positive integers are

→ x , (x+1) , (x+2) , (x+3) and (x+4)

So, as said in question average of the squares of five consecutive positive integers is 66

→ x² + (x+1)² + (x+2)² + (x+3)² + (x+4)² = 66

By solving above equation we get :-

= $\sf{\frac{5x^{2} + 20x + 30 }{5} - 66 = 0}$

= $\sf{\frac{5(x^{2}+4x+6)}{5} - 66 = 0}$

= x² + 4x + 6 - 66 = 0

= x² + 4x - 60 = 0

Now by using middle term splitting method

= x² + 10x - 6x - 60 = 0

= x( x + 10 ) - 6( x + 10 ) = 0

= ( x + 10 )( x - 6 ) = 0

→ x + 10 = 0

→ x = - 10 [neglect]

→ x - 6 = 0

→ x = 6

We consider the positive value of " x " as said in question

So , the five consecutive numbers are :-

→ x , (x+1) , (x+2) , (x+3) and (x+4)

Now by substituting value of " x " :-

→ 6, (6+1) , (6+2) , (6+3) and (6+4)

→ 6 , 7 , 8 , 9 and 10

Now we find the average of these numbers

• no. of terms = 5
• sum of all terms = 40

→ 6 + 7 + 8 + 9 + 10 = 40

= $\sf{\frac{6+7+8+9+10}{5}}$

= $\sf{\cancel\frac{40}{5}}$

= 8

Hence the average of these five integers is 8