Question

Question No 9
Find sum of three consecutive positive integers whose product
is equal to sixteen times their sum.
23
20
22
21​

Answer 1

Answer:

21

Step-by-step explanation:

let the consecutive positive integers be = a-1,a,a+1

given,

the product is equal to sixteen times their sum

(a-1)a(a+1)=16(a-1+a+a+1)

a($a^{2} -1^{2}$) =16(3a)

a($a^{2} -1^{2}$) =48a

($a^{2} -1^{2}$)=48

$a^{2}$=49

a=7

not taking -7 as it is given in the question that the three numbers are positive integers

therefore the numbers will be 6,7,8

the sum of them will be 6+7+8=21