Question

The sum of three numbers in an Arithmetic Progression is 3 & their product is -35,
find the numbers.

Answer 1

$\large\bf\underline {To \: find:-}$

• Three numbers of AP.

$\large\bf\underline{Given:-}$

• sum three numbers = 3
• product of three numbers = -35

$\huge\bf\underline{Solution:-}$

Let the three numbers be (a - d) , a , (a + d)

❥ According to Question

Sum of three numbers in a AP = 3

➛ (a - d) + a + (a + d) = 3

➛ a + a + a - d + d = 3

➛ 3a = 3

➛ a = 3/3

➛ a = 1 .......1)

Product of three numbers = -35

➛(a - d) × a × (a + d) = -35

➛ a(a² - d²) = - 35

➛ a³ - ad² = -35

Substituting value of a from 1)

➛ 1³ - 1 × d² = -35

➛ 1 - d² = -35

➛ - d² = -35 - 1

➛ - d² = - 36

➛ d = √36

➛ d = 6

So,

⏺️three numbers in an AP are :-

a - d = 1 - 6 = - 5

a = 1

a + d = 1 + 6 = 7

hence,

three numbers in an AP are -5 , 1 ,and 7

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Answer 2

Answer:

Solution :-

Let the number are x−y,x,x+y

Sum =−3

⇒x−y+3x+x+y=−3

⇒3x=−3

⇒x=−1

Now product =8

⇒(x−y)(x)(x+y)=8

Substituting x=−1

we get (−1−y)(−1)(−1+y)=8

(y^2 −1)=8

⇒y=±3

The no : are −4,−1,2 or 2,−1,−4  

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