Question

Sum of three terms of an A.P. is 57 and their product is 6840 find the smallest number of the A.P.​

Answer 1

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.

Answer 2

Step-by-step explanation:

Let the three numbers be

$a + d \\ a - d \\ a$

now ,

a+ d +a -d +a =57

3a = 57

a= 19

now,

$(a + d)(a - d)a = 6840 \\ (19 + d)(19 - d)19 \\ using \: a {}^{2} - {b}^{2}$

$(361 - {d}^{2} )19 = 6840 \\ 361 - {d}^{2} = \frac{6840}{19} \\ - {d}^{2} = - 1 \\ d = 1$

now the smallest number will be the first term i.e. a =19

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