Question

17. There are three consecutive positive
integers, such that the sum of the squares
of the first and the product of the other two
is 154. Find the integer which is a multiple
of 3?​

Answer 1

Given:

• There are three consecutive positive integers, such that the sum of the squares of the first and the product of the other two is 154.

To find:

• Find the integer which is a multiple of 3?​

Now,

Let the three consecutive number be (n), (n+1), and (n+2)

So,

• According to the question, the equation which is formed is

⇒ n² + (n+1) * (n+2) = 154

⇒ n² + n(n+2) + 1(n+2) = 154

⇒ n² + n² + 2n + n + 2 = 154

⇒ n² + n² + 3n + 2 = 154

⇒ 2n² + 3n = 154 - 2

⇒ 2n² + 3n = 152

⇒ 2n² + 19n - 16n - 152 = 0

⇒ n(2n + 19) - 8(2n + 19) = 0

⇒ (n - 8)(2n + 19) = 0

So,

(n - 8) = 0, and (2n + 19) = 0

n = 8, and n = (-19/2)

Thus,

• n = (-19/2) this does not belong to integers.

• So we accept n = 8.

Now,

The three consecutive positive integers are

• n = 8
• (n + 1) = (8 + 1) = 9
• (n + 2) = (8 + 2) = 10

So,

• From 8, 9, and 10 only 9 is multiple of 3.

Also, we can say that (n+1) = 9 is the multiple of 3

Answer 2

GIVEN :-

• Sum of the squares of the first and the product of the other two is 154

TO FIND :-

• The one of the integer which is divisible by 3

SOLUTION :-

Let the three consecutive positive integers be ,

• a , a + 1 , a + 2

Now ,

Square of the first positive integer = a²

Product of the second and third positive integers =

(a +1 )( a + 2)

Now According to given Condition ,

$\tt : \implies \: {a}^{2} + (a + 1)(a + 2) = 154 \\ \\ \tt : \implies \: {a}^{2} + {a}^{2} + 2a + a + 2 = 154 \\ \\ \tt : \implies \: 2 {a}^{2} + 3a + 2 = 154 \\ \\ \tt : \implies \: 2 {a}^{2} + 3a = 154 - 2 \\ \\ \tt : \implies 2 {a}^{2} + 3a - 152 = 0 \\ \\ \red \bigstar \sf \: by \:Spiliting \: the \: middle \: term \: method , \: \\ \\ \tt: \implies2 {a}^{2} - 16a +19a - 152 = 0$

$\tt : \implies \: 2a(a - 8) + 19(a - 8) = 0 \\ \\ \red \bigstar \sf \: Taking \: (a - 8) \:as\: common \: we \: get ,\: \\ \\ \tt : \implies (a - 8)(2a + 19) = 0 \\ \\ \tt : \implies \: a = 8 \: (or) \: a = - \frac{19}{2}$

According to the condition mentioned in question ' the three consecutive positive integers ' . a can take only positive values So ,

$\tt : \large \: \implies {\underline {\boxed {\bf{a = 8}}}} \: \blue\bigstar$

Then the remaining consecutive positive integers are ,

$\bf \: a + 1 = 8 + 1 = 9 \\ \\ \bf \: a + 2 = 8 + 2 = 10$

Hence , The three positive consecutive Integers satisfying the given condition are 8 , 9 and 10

Among these three consecutive positive integers , only 9 is multiple by 3 .

∴ The integer among these three consecutive positive integers satisfying the given condition as well as the multiple of 3 is 9