Question

find three consecutive terms in A.P. whose sum is9 and the product of their cubes is 3375​

Answer 1

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

Sum = a - d + a + a + d = 3a = 9

Thus the middle term, a = 3.

Given,

[(a - d) a (a + d)]³ = 3375

(a - d) 3 (a + d) = 15

a² - d² = 5

d² = 3² - 5

d² = 4

d = ± 2

Well we get the same three terms whenever d = 2 or d = -2.

So, the terms are 1, 3 and 5.

Answer 2

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Given:

The consecutive terms in A.P. whose sum is 9 and the product of their cubes is 3375.

To find:

The three consecutive terms in A.P.

Explanation:

Let the three consecutive terms in an A.P. be a-d, a, a+d.

Given,

The sum of these three consecutive numbers is 9.

→ a - d + a + a + d = 9

→ 3a = 9

→ a = $\cancel{\frac{9}{3} }$

→ a = 3

The product of their cube is 3375.

⇒ (a-d)³ × a³ × (a+d)³ = 3375

Putting the value of a in this product,we get;

⇒ (3-d)³ × (3)³ × (3+d)³ = 3375

⇒ (3-d)³ × 27 × (3+d)³ = 3375

⇒ (3-d)³ × (3+d)³ = $\cancel{\frac{3375}{27} }$

⇒ (9+3d -3d -d²)³ = 125

⇒ (9 -d²)³ = (5)³

⇒ $(9-d^{2} )^\cancel{{3} }\:=(5)^\cancel{{3} }$

⇒ 9 - d² = 5

⇒ -d² = 5 - 9

⇒ -d² = -4

⇒ d² = 4

⇒ d= √4

⇒ d = 2

∴ Three consecutive terms in Arithmetic Progression;

• 1st term, a-d = 3 -2= 1
• 2nd term, a= 3
• 3rd term, a+d= 3+2= 5