Question

The average of four consecutive odd numbers is 56. What is the product of first and last term?

Answer 1

Given :

• Average of four consecutive odd numbers = 56

To find :

• Product of first and last term

Solution :

Let us assume the four consecutive odd numbers as x, x + 2, x + 4, x + 6

• First consecutive odd number = x
• Second consecutive odd number = x + 2
• Third consecutive odd number = x + 4
• Fourth consecutive odd number = x + 6

Remember : Whenever we have to assume odd or even numbers we assume it as x, x +2, x + 4... that is by assuming them as any variable and adding consecutive even number to it. And whenever we assume natural numbers, whole numbers, integers, etc we assume them as x, x + 1, x + 2,... that is assuming them as any variable and adding a consecutive number to it.

The average of x, x + 2, x + 4, x + 6 = 56

Formula of average :-

$\boxed{ \bf{ \blue{Average = \dfrac{Sum \: of \: all \: the \: items}{Total \: number \: of \: items}}}}$

→ Sum of all the consecutive odd numbers = x + x + 2 + x + 4 + x + 6

→ Sum of all the consecutive odd numbers = 4x + 12

→ Total number of consecutive odd numbers = 4

Substituting the values :-

$\implies \sf Average = \dfrac{Sum \: of \: all \: the \: items}{Total \: number \: of \: items}$

$\implies \sf 56 = \dfrac{4x + 6}{4}$

Taking 2 common from the numerator :-

$\implies \sf 56 = \dfrac{2(x + 3)}{4}$

$\implies \sf 56 = \dfrac{\not 2(x + 3)}{\not4}$

$\implies \sf 56 = \dfrac{x + 3}{2}$

$\implies \sf 56 \times 2 = x + 3$

$\implies \sf 112 = x + 3$

$\implies \sf 112 - 3 = x$

$\implies \sf 109 = x$

→ The value of x = 109

Substitute the value of x in the four consecutive odd numbers :-

→ First consecutive odd number = x = 109

→ Second consecutive odd number = x + 2 = 109 + 2 = 111

→ Third consecutive odd number = x + 4 = 109 + 4 = 113

→ Fourth consecutive odd number = x + 6 = 109 + 6 = 115

Product of the first and last term :-

→ First term × Fourth term

→ 109 × 115

→ 12535

Therefore, the product of the first and last term = 12535