Question

In an A.P the sum of three consecutive terms is 57 & their product is 6555. Find the A.P.​

Answer 1

Answer:

Step-by-step explanation:

Answer:

The smallest number of the A.P. is 18.

Given:

Sum of three terms of an A.P. is 57 and their product is 6840.

To find:

Smallest number of the A.P.

Solution:

Let the three term of the A.P. be (a-d), a, (a+d) respectively .

According to the first condition

>> (a-d)+a+(a+d)=57

>> 3a=57

>> a=57/3

>> a=19

According to the second condition.

>> (a-d) (a) (a+d)=6840

>> (a²-d²) (a)=6840

Substitute a=19

>> (19²-d²) (19)=6840

>> 361-d²=6840/19

>> d²=361-360

>> d²=1

>> d=1 or -1

If d=1,

a-1=19-1=18,

a=19,

a+1=19+1=20

If d=-1,

a-1=19-(-1)=20,

a=19,

a+1=19+(-1)=18

The smallest number of the A.P. is 18.