Question

the sum of three consecutive odd number is 63. Find the number which is 18 more than the second number. ​

Answer 1

Answer:

Your answer: 39

Explanation:

let First number be x

(x)+(x+2)+(x+4)=63

= 3x+6=63

= 3(x+2)=21

= x+2= 21-2

= x= 19

Now second number is x+2

=19+2

=21

As we need 18 more than second number

Hence, answer is 18+21= 39

Mark brainliest❤

Answer 2

Let the three odd consecutive number be:

1. ⇒a - d
2. ⇒a
3. ⇒a + d

Now,

According To Question,

$\tt \implies a - d + a + a + d = 63 \\ \tt \implies 3a = 63 \\ \tt \implies \: a = \frac{63}{3} \\ \tt \implies \: a = 21$

Hence,

✏ 2nd number = 21

We have to find the number which is 18 more than the second number.

Then, The number will be:

$\large \implies {\boxed {\tt \blue { \: = (21 + 18) => 39 }}}$