Question

If the sum of three consecutive
odd numbers is 45, then find the
largest of these numbers.​

Answer 1

Step-by-step explanation:

sum is 3 consecutive odd numbers is..

1st no. =x

2nd no.=x+3

3rd no.=x+5

sum of x+x+3+x+5=45

= 3x+8=45

= 3x=45-8,=36

= 3x =36

x=12

x=12, x+3=15 ,x+5=17

greatest number is 17

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Answer 2

$\rule{400}4$

☆ ANSWER:-

▪︎ Given:-

• Sum of three consecutive odd integers = 45.

▪︎ To Find:-

• The largest of the number.

$\rule{400}2$

▪︎ Now,

• These numbers are consecutive odd integers.

• So let the integers be (x), (x+2), (x+4).

$\rule{400}2$

▪︎ Therefore ATQ:-

$\tt\mapsto(x) + (x + 2) + (x + 4) = 45. \\ \\ \tt\mapsto x + x + x + 2 + 4 = 45. \\ \\ \tt\mapsto3x + 6 = 45. \\ \\ \tt\mapsto3x = 45 - 6. \\ \\ \tt\mapsto3x = 39. \\ \\ \tt\mapsto x = \cancel \frac{39}{3}. \\ \\ \tt\mapsto x = 13.$

$\rule{400}2$

▪︎ So,

• The numbers are:-

• x = 13.

• x+2 = 13+2 = 15.

• x+4 = 13+4 = 17.

▪︎ Thus the largest of the number is 17.

$\rule{400}4$