Question

There are five consecutive integers.Thrice the middle integer exceeds the sum of the greatest and the least integers by 12.Find the greatest integer​

Answer 1

Answer:-

The Greatest integer is 14.

Explanation:-

Given:-

• There are five consecutive integers.

• Thrice of middle integer exceeds sum of the greatest and least integer be 12.

To Find:-

• The Greatest integer.

Now,

Since these are 5 consecutive integers so let the integers be (x), (x+1), (x+2), (x+3) and (x+4) Respectively.

Clearly,

• Middle integer is (x + 2).

• Least integer is (x).

• Greatest integer is (x + 4)

Therefore ATQ:-

Three times the middle term = 12 + (least integer + greatest integer).

$\\ \tt\mapsto3( x + 2) = 12 + \bigg(x + (x + 4) \bigg).$

$\\ \tt\mapsto3x + 6 = 12 + \bigg(2x + 4 \bigg).$

$\\ \tt\mapsto 3x + 6 = 16 + 2x.$

$\\ \tt\mapsto3x - 2x = 16 - 6.$

$\\ \large\red{\mapsto \boxed{ \blue{ \bf x = 10.}}}$

So The value of x is 10.

Amd Hence,

Greatest Integer,

= x + 4.

= 10 + 4.

= 14.

$\bf\therefore$ The Required Greatest Integer is 14.

Answer 2

Step-by-step explanation:

There are five consecutive integers.. Assume that as x, (x + 1), (x + 2), (x + 3) and (x + 4).

Thrice the middle integer exceeds the sum of the greatest and the least integers by 12.

As per given condition,

→ 3(x + 2) = (x + 4) + x + 12

→ 3x + 6 = x + 4 + x + 12

→ 3x + 6 = 2x + 16

→ 3x - 2x = 16 - 6

→ x = 10

Therefore,

• x = 10
• x + 1 = 10 + 1 = 11
• x + 2 = 10 + 2 = 12
• x + 3 = 10 + 3 = 13
• x + 4 = 10 + 4 = 14

In the assumed consecutive integers x is the smallest one and (x + 2) is the middle one while the (x + 4) is the greatest one.

We have to find thegreatest integer.

Hence, (x + 4) i.e. 14 is the greatest integer.