Question

The sum of 3 consecutive integers is 66.
What is the third number in this sequence?

Answer 1

Answer :

Let the first number be x

Then the numbers are, x, x+1, x+2

Sum = 66

x+x+1+x+2 = 66

3x + 3 = 66

3x = 66-3

3x = 63

x = 63/3

x = 21

First no. = x = 21

Second no. = x+1 = 21+1 = 22

Third no. = x+2 = 21+2 = 23

The numbers are 21, 22, 23

Hope it helps...

Answer 2

Answer:-

$\red{\bigstar}$ The 3rd number in sequence will be $\large\leadsto{\sf\green{23}}$

• Given:-

Sum of 3 consecutive integers = 66

• To Find:-

3rd number in the sequence = ?

• Solution:-

Let the 3 consecutive numbers be x, x+1, x+2.

ACCORDING TO QUESTION:-

$\implies\sf{x + (x+1) + (x+2) = 66}$

$\implies\sf{3x + 3 = 66}$

$\implies\sf{3(x+1) = 66}$

$\implies\sf{x+1 = \dfrac{66}{3}}$

$\implies\sf{x+1 = 22}$

$\implies\sf{x = 22 - 1}$

$\implies\bf\red{x = 21}$

Hence, the 3 numbers are:-

$\orange{\bigstar}$ x = 21

$\blue{\bigstar}$ x+1 = 21 + 1 → 22

$\green{\bigstar}$ x+2 = 21 + 2 → 23

Therefore, the 3rd number in the. sequence will be $\large\bf\red{23}$