Question

The average of 5 consecutive even numbers a, b, c, d and e is 106. what is the product of b and d ?

Answer 1

0is the answer of this question.

Answer 2

Answer:

Product of b and d = 104 × 108 = 11,232

Step-by-step explanation:

Given: a, b, c ,d and e are consecutive even numbers

and we know consecutive even numbers are of the form 2n , 2n + 2, 2n+4 .........and so on

Thus,

a = 2n

b = 2n + 2

c = 2n + 4

d = 2n + 6

e = 2n + 8

Average is sum of number divided by number of terms

$\frac{a+b+c+d+e}{5}=106$

Substitute, the above values ,

$\frac{2n+2n+2+2n+4+2n+6+2n+8}{5}=106$

$\frac{10n+20}{5}=106$

$2n+4=106$

$2n=106-4=102$

$n=51$

thus, b = 2n + 2 = 104

d = 2n + 6 = 108

Thus, Product of b and d = 104 × 108 = 11,232