Question

find three consecutive odd integers such as the sum of forst and second integer exceed the third integer by 19​ .Now please answer tis question correctly please​

Answer 1

Answer:

21, 23, 25

Step-by-step explanation:

Consecutive odd number come at a difference of 2.

Let those numbers are: a - 1,

a - 1 + 2 = a + 1 and

a + 1 + 2 = a + 3.

In question,

= > (a - 1) + (a + 1) = (a + 3) + 19

= > a - 1 + a + 1 = a + 3 + 19

= > a + a = a + 22

= > 2a = a + 22

= > 2a - a = 22

= > a = 22

Hence, numbers are:

a - 1 = 22 - 1 = 21

a + 1 = 22 + 1 = 23

a + 3 = 22 + 1 = 25

Answer 2

Correct question:

Find the three consecutive odd integers such as the sum of first and second integer exceed the third integer by 19.

_______________________________________

Answer:

$\sf{The \ three \ consecutive \ odd \ integers \ are}$

$\sf{21, \ 23 \ and \ 25.}$

Given:

$\sf{In \ three \ consecutive \ odd \ integers,}$

$\sf{sum \ of \ first \ and \ second \ integer \ exceed}$

$\sf{the \ third \ integer \ by \ 19.}$

To find:

$\sf{The \ three \ consecutive \ odd \ integers.}$

Solution:

$\sf{Let \ the \ three \ consecutive \ odd \ integers \ be}$

$\sf{(n-2), \ n, \ and \ (n+2).}$

$\sf{According \ to \ the \ given \ condition.}$

$\sf{(n-2)+n=(n+2)+19}$

$\sf{\therefore{2n-2=n+21}}$

$\sf{\therefore{2n-n=21+2}}$

$\sf{\therefore{n=23}}$

$\sf{The \ three \ consecutive \ odd \ integers \ are}$

$\sf{n-2=23-2=21,}$

$\sf{n=23,}$

$\sf{n+2=25}$

$\sf\purple{\tt{\therefore{The \ three \ consecutive \ odd \ integers \ are}}}$

$\sf\purple{\tt{21, \ 23 \ and \ 25.}}$