Math, asked by ravinarula10, 11 months ago

Find three consecutive odd integers such that the sum of the
first, two times the second, and three times the third is 70.

Answers

Answered by kldhingra38
3

Answer:

9,11,13

Step-by-step explanation:

Given :The sum of the first, twice the second, and three times the third = 70

To find : The three consecutive odd integers.

Proof:

Let the first odd integer = x ,

The second consecutive odd integer = x+2 ,

The third consecutive odd integer = (x+2)+2 =  x+4

So,The sum of the first, twice the second, and three times the third will be

x+2(x+2)+3(x+4)

Given the sum of the first, twice the second, and three times the third = 70

∴   x+2(x+2)+3(x+4) = 70

=> x+2x+4+3x+12 = 70

=> 6x+16 = 70

=> 6x = 70-16

=> 6x = 54

=> x = 54÷6

=> x = 9

Now we have;

first odd integer = x

=> x= 9

The second consecutive odd integer = x+2

=> x+2

=> 9+2 = 11

=> x = 11

The third consecutive odd integer = x+4

=> x+4

=> 9+4 = 13

=> x = 13

∴ the three consecutive odd integers x=9, x=11, x= 13.

Hence, the three consecutive odd integers are 9,11,and 13

Answered by Sike15
1

Answer:

9, 11, 13

Step-by-step explanation:

SOLUTION:

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