Question

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

Answer 1

Here is the answer of your question

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

Answer:

39

Step-by-step explanation:

Let the first odd number = x

second odd number = x + 2

and third odd number = x + 4

According to the condition,

x+ x + 2 + x+4 = 63

3x + 6 = 63 ⇒ 3x = 63 – 6

⇒3x = 57 ⇒ x = \frac { 57 }{ 3 } =19

∴ First odd number = 19

Second odd number = 19 + 2 = 21

third odd number = 19 + 4 = 23

required number which is 18 more than the second number is 18+21=39

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Answer 2

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\:\:$

$\green{\underline \bold{Given :}}$

$\:\:$

• Sum of consecutive 3 odd number is 63

$\:\:$

$\red{\underline \bold{To \: Find:}}$

$\:\:$

• Number which is 18 more than the second number.

$\:\:$

$\large{\orange{\underline{\tt{Solution :-}}}}$

$\:\:$

Let the first odd number be 'a'

Hence its successive consecutive odd number will be 'x+2'

Again its successive consecutive odd number will be 'x+4'

$\:\:$

$\purple{\underline \bold{According \: to \: the \ question :}}$

$\:\:$

$\sf \implies x + (x + 2) + (x + 4) = 63$

$\:\:$

$\sf \implies 3x + 6 = 63$

$\:\:$

$\sf \implies 3x = 57$

$\:\:$

$\sf \implies x = \dfrac { 57 } { 3 }$

$\:\:$

$\sf \implies x = 19$

$\:\:$

Hence,

$\:\:$

1. First number = 19
2. Second number = 19 + 2 = 21
3. Third number = 19 + 4 = 23

$\:\:$

Second number = 21

$\:\:$

$\footnotesize{ \underline{\bold{\texttt{Number which is 18 more than 2nd number :}}}}$

$\:\:$

$\sf \implies 21 + 18$

$\:\:$

$\sf \implies \boxed{39}$

$\rule{200}5$