Question

find three consecutive terms in AP whose sum is 9 and the product of their cube is 33 75 ​

Answer 1

Answer:

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.

Answer 2

Answer:

let the 1st no. be x

let the 2nd no. be x+1

let the 3rd no. be x+2

x+x+1+x+2=9

3x+3=9

x=2

x+1=2+1

=3

x+2=2+2

=4

2³*3³*4³

=8*27*64

=3375