Question

V) Find three consecutive terms in A.P. whose sum is 9 and the product of their
cubes is 3375.​

Answer 1

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

Step-by-step explanation:

Let the three consecutive terms in A.P. be “(a-d)”, “a” & “(a+d)”.

So, the sum of the three consecutive terms in A.P. = (a-d) + a + (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 = 3 ……. (i)

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 from (i)]

⇒ [(3-d) * (3+d)]³ = 3375 / 27  

⇒ [(3-d) * (3+d)] = ∛125

⇒ 3² – d² = 5

⇒ d² = 9 – 5 = 4

⇒ d = 2 …… (ii)

Therefore, from (i) & (ii), we get

The three consecutive terms in A.P. are,

a-d = 3 – 2 = 1

a = 3

a+d = 3+2 = 5

Thus, the three consecutive terms in A.P. are 1, 3 & 5.

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Answer 2

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