Question

find three consecutive terms oa AP whose sum ia 9and product of their cubes is3375​

Answer 1

Answer:

The three consecutive terms in A.P. is 1, 3 & 5.

Step-by-step explanation:

Let the three consecutive terms in A.P. be

First term=(a-d)

Second term=a

Third term=(a+d)

But, Sum of the 3 consecutive terms is given = 9

⇒ (a-d) + a + (a+d) = 9

⇒ 3a – d + d = 9

⇒ 3a = 9  

⇒ a = 9/3

a= 3

Also given, the product of their cubes = 3375

∴ (a-d)³ * a³ * (a+d)³ = 3375

⇒ (3-d)³ * 3³ * (3+d)³ = 3375                        [∵ a = 3]

⇒ [(3-d) * (3+d)]³ = 3375 / 27  

⇒ [(3-d) * (3+d)] = ∛125

⇒ 3² – d² = 5

⇒ d² = 9 – 5 = 4

⇒ d = 2

The three consecutive terms in A.P. are,  

First term=a-d

= 3 – 2

= 1

Second term=a = 3

Third term=a+d

= 3+2

= 5

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