Question

Find three consecutive terms of an arithmetic progression whose sum is 27 and their product is 504

Answer 1

no=(a-d),a,a+d

a-d+a+a+d=27

3a=27

a=9

(9-d)(9)(9+d)=504

(81-d²)(9)=504

729-9d²=504

9d²=225

d²=225/9

d²=25

d=+-5

1st=5,14

2nd=9

3rd=14,5

Answer 2

Answer:

Let three consecutive terms are

a - d, a, a + d

$(a - d) + a + (a + d) = 27$

$a - d + a + a + d = 27$

$\: \: \: \: \: \: a + a + a = 27$

$\: \: \: \: \: \: 3a = 27$

a = 27/3

$\: \: \: \: \: \: \: a = 9$

$(a - d) \times a \times (a + d) = 504$

$(9 - d)(9 + d) = 504 \div 9$

$\: \: \: \: \: \: \: \: \: \: \: (a + b)(a - b)$

$\: \: \: \: \: \: \: = a {}^{2} - b {}^{2}$

$9 {}^{2} - d {}^{2} = 56$

$81 - d {}^{2} = 56$

$- d {}^{2} = 56 - 81$

$- d {}^{2} = - 25$

$d = \sqrt{25}$

$d = 5$

$d = 5 \: or \: d = - 5$

A.P. when a = 9, d = 5.

a - d, a, a + d

9 - 5, 9, 9 + 5, ....

4, 9, 14, ....

A.P. when a = 9, d = - 5

a - d, a, a + d, .....

9 - (- 5), 9, 9 + (- 5)

9 + 5, 9, 9 - 5

14, 9, 4, ....

∴ The three consecutive terms are 4,9,14 Or 14,9,4.

Step-by-step explanation:

@Genius