Question

are the sum of the three consecutive terms of an ap is 30 and their product is 360 find the terms​

Answer 1

Answer:

Step-by-step explanation:

Let the three consecutive no. be

a-d,a ,a+d.

Therefore a-d+a+a+d=30.

3a= 30

a=10.

(a-d)(a)(a+d)=360.

Solve substituting a=10

You will get d=+-8

Answer 2

Step-by-step explanation:Answer:

Step-by-step explanation:Let the three consecutive terms of the A.P be (a-d),(a) , (a+d).

From First condition,

(a-d)+(a)+(a+d)=30

a+a+a=30

3a=30

a=10

From 2nd condition,

(a-d)×(a)×(a+d)=360

Putting a=10,

(10-d)×(10)×(10+d)= 360

(10-d)×(10+d)= 360/10

(10-d)×(10+d)= 36

Bracket multiplication,

100-d²=36

d²=100-36

d²=64

d= +_8.

The required three consecutive terms are:

1.(a-d) =10-8=2

2.(a)= 10

3.(a+d)= 10+8=18

OR

1.(a-d) =10-(-8)=18

2.(a)= 10

3.(a+d)= 10+(-8)=2

Ans: The three consecutive terms are 2,10,18 OR 18,10,2.

Hope it helps!!!