Question

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

Answer 1

$\huge\star{\green{\underline{\mathfrak{Answer :-}}}}$

Terms :- 1,3, 5.

$\huge\star{\green{\underline{\mathfrak{Explanation :-}}}}$

Let the three consecutive terms of the AP be a,a+d and a-d.

It is given that,

$a + a + d + a - d = 9$

$3a = 9$

$a = 3$

It is also given that,

${a}^{3} + {(a + d)}^{3} + {(a - d)}^{3} = 3375$

${3}^{3} + {(3 + d)}^{3} + {(3 - d)}^{3} = 3375$

$27 + {(3 + d)}^{3} + {(3 - d)}^{3} = 3375$

$({3 + d)}^{3} {(3 - d)}^{3} = \dfrac{3375}{27}$

${((3 + d)(3 - d))}^{3} = 125$

$(3 + d)(3 - d) = \sqrt[3]{125}$

$(3 + d)(3 - d) = 5$

${3}^{2} - {d}^{2} = 5$

${d}^{2} = 9 - 5 = 4$

$d = 2$

So, the terms of the AP are : 3, 3+2=5 and 3-2=1.

______________