Question

find three consecutive terms in an A.P.whose sum is 9 and product of their cube is 3357​

Answer 1

The first three consecutive terms in an A.P. are "(1, 3, 5) or (5, 3, 1)".

Step-by-step explanation:

Let (a -d), a and (a + d) be the three consecutive ters in A.P.

According to question,

(a -d) + a + (a + d) = 9

⇒ 3a = 9

⇒ a = 3

Also,

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

⇒ ${ (a -d).a.(a + d)}^{3}$  =$15^{2}$

⇒ (a - d).a.(a + d) = 15

Put a =3, we get

9 - $d^{2}$ = $\frac{15}{3}$ = 5

⇒ $d^{2}$ = 9 - 5 = 4

⇒ d = ± 2

∴ a - d = 3 - 2 = 1 and  a+ d = 3 + 2 = 5 ( d = 2)

∴ a - d = 3 + 2 = 5 and  a + d = 3 -2 = 1 ( d = - 2)

The first three consecutive terms in an A.P. are "(1, 3, 5) or (5, 3, 1)".

Answer 2

Given: Find three consecutive terms in an A.P. whose sum is $9$ and product of their cube is $3357$​

To Find: The three consecutive terms in A.P.

Step-by-step explanation:

Let the three consecutive terms in an A.P. be,

$a-d,a,a+d$

From the given statements,

The sum of these three consecutive numbers is $9$.

• $(a-d)+a+(a+d)=9$

⇒$a-d+a+a+d=9$

⇒$3a=9$

∴ $a=3$

And The product of their cube is $3357$.​

• $(a-d)^{3}\times a^{3} \times (a+d)^{3}=3357$

Plug the value of $a=3$ ,

⇒$(3-d)^{3}\times 3^{3} \times (3+d)^{3}=3357$

⇒ $(3-d)^{3}\times (3+d)^{3}=\frac{3357}{3^{3} }$

⇒$(9+3d -3d -d^{2} )^{3} = 125$

⇒$(9 -d^{2} )^{3} = (5)^{3}$

Cancel cube on both sides,

⇒$9 -d^{2} = 5$

⇒$d^{2}=9-5=4$

⇒$d=\pm 2$

∴ Three consecutive terms in Arithmetic Progression,

→ When $a=3\ and\ d=2$

Then,

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

→When $a=3\ and\ d=-2$

Then,

• 1st term, $a-d=3-(-2)=5$

• 2nd term, $a=3$

• And 3rd term $a+d=3-2=1$