Verified answer

Step-by-step explanation:

Let one side be a, and the common difference be d,

Then sides are a, a+d, a+2d …………… Arithmatic Sequence

Given consition: a + a+d + a+2d = 30

OR 3a + 3d = 30

OR a + d = 10

Possible acceptable vales of the three sides are:

6,10,14

7,10,13

8,10,12,

9,10,11

Answer:

Let the sides of triangle is a-d ,a,a+d where a,d>0

Let the sides of triangle is a-d ,a,a+d where a,d>0=> a-d+a+a+d=30

=> 3a=30→a=10

Properties of triangle:—

Properties of triangle:—Sum of any two sides of a triangle

Properties of triangle:—Sum of any two sides of a triangleis always greater than the third side.

always greater than the third side.=> a + a-d > a+d

always greater than the third side.=> a + a-d > a+d=> a>2d

always greater than the third side.=> a + a-d > a+d=> a>2d=> 5>d (since a=10)

always greater than the third side.=> a + a-d > a+d=> a>2d=> 5>d (since a=10)So, all possible sides of triangle are

9,10,11

9,10,118,10,12

9,10,118,10,127,10,13

Step-by-step explanation:

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