Question

In an A.P. sum of three consecutive terms is 27 and their products
is 504. Find the terms.​

Answer 1

Step-by-step explanation:

let the terms be a-d, a and a+d.

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

simplifying we get,

$3a = 27$

$a = 9$

thus the terms are

9-d, 9, 9+d

i hope it helps.

Answer 2

Answer:

• Three consecutive terms are 4, 9 and 14.
• or, 14, 9 and 4.

Step-by-step explanation:

Given that:

In an A.P.

• Sum of three consecutive terms is 27.
• Their products is 504.

To Find:

• The terms.

Let us assume:

• First term be (a - d).
• Second term be a.
• Third term be (a + d).

According to the question.

Sum of consecutive terms = 27

⟶ (a - d) + a + (a + d) = 27

⟶ a - d + a + a + d = 27

Cancelling d.

⟶ 3a = 27

⟶ a = 27/3

⟶ a = 9

Products of consecutive terms = 504

⟶ (a - d) × a × (a + d) = 504

Substituting the value of a.

⟶ (9 - d) × 9 × (9 + d) = 504

⟶ (9 - d) × (9 + d) = 504/9

Using (a - b) (a + b) = a² - b²

⟶ 9² - d² = 56

⟶ 81 - d² = 56

⟶ d² = 81 - 56

⟶ d² = 25

⟶ d = √25

⟶ d = ± 5

Three consecutive terms are:

When d = 5

• First term = (9 - 5) = 4
• Second term = 9
• Third term = (9 + 5) = 14

When d = - 5

• First term = 9 - (-5) = 9 + 5 = 14
• Second term = 9
• Third term = 9 + (- 5) = 9 - 5 = 4