Question

The sum of three numbers in A.P is -3 and the product is 8 then find the three numbers

Answer 1

GIVEN :-

Sum of three numbers in A.P is -3 and the product is 8.

• S = -3

• P = 8

TO FIND :-

The three numbers which are in A.P.

SOLUTION :-

Let the three numbers be a - d, a and a + d.

A/q,

S = a - d + a + a + d = -3

⇒a + a + a = -3

⇒3a = -3

⇒a = -3/3

⇒a = -1

$\rule{110}2$

Again,

P = a(a - d)(a + d) = 8

Putting the value of a :-

⇒ -1(-1 - d)(-1 + d) = 8

⇒ -1 × [ (-1)² - (d)²] = 8

⇒ -1 × (1 - d²) = 8

⇒ -1 + d² = 8

⇒ d² = 9

⇒d = √9

⇒d = ±3

$\rule{110}2$

CASE - 1 :-

If d = +3, the numbers are :-

• a - d = -1 - 3 = -4

• a = -1

• a + d = -1 + 3 = 2

$\rule{80}2$

CASE - 2 :-

If d = -3, the numbers are :-

• a - d = -1 - (-3) = -1 + 3 = 2

• a = -1

• a + d = -1 + (-3) = -1 - 3 = -4

Therefore, the numbers are 2, -1 and -4.

Answer 2

$\rule{200}3$

$\huge\tt{SOLUTION:}$

• Sum of 3 numbers in A.P. is -3
• Product of them is 8

$\rule{200}3$

$\huge\tt{TO~FIND:}$

• Those three numbers

$\rule{200}3$

$\huge\tt{SOLUTION:}$

Let those numbers be a - d, a and a + d.

A/Q,

➠S = a - d + a + a + d = - 3

➠a + a + a = - 3

➠3a = -3

➠a = -3/3

➠a = -1 (cancelling 3 by 3)

➠a = -1 ____(EQ.1)

$\rule{200}3$

Then,

➠P = a(a-d)(a+d) = 8

So, if we put value from (EQ.1),

➠-1(-1-d) (-1+d) = 8

➠-1 × [(-1)² - (d)²] = 8

➠-1 × (1-d²) = 8

➠-1 + d² = 8

➠d² = 8 + 1

➠d² = 9

➠d = √9 or ±3

$\rule{200}3$

So, by assuming these there can be two situations,

If D = +3 , the numbers would be

⟹a - d = -1 - 3 = 1+3 = 4

⟹a + d = -1 + 3 = 2

⟹a = -1

And if D = -3, the numbers would be

⟹a - d = -1-(-3) = -1 + 3 = 2

⟹a + d = -1 + (-3) = -1 -3 = -4

⟹a = -1

Hence, The Numbers Are 2 , -1 , -4

$\rule{200}3$