Question

9. Find three consecutive odd numbers whose sum is 147.
Hint. Let the required numbers be (2x+1), (2x+3) and (2x+5).​

Answer 1

Answer:

The three consecutive odd numbers are as follows :-

• $\sf {1^{st}}$ number = 47
• $\sf {2^{nd}}$ number = 49
• $\sf {3^{rd}}$ number = 51

Step-by-step explanation:

In this question, we are given with the sum of three consecutive odd numbers that is 147.

We've been asked to find out the three consecutive odd numbers. How can we find? In order to find numbers, let us assume three consecutive numbers as follows :-

• $\sf {1^{st}}$ number = 2x + 1
• $\sf {2^{nd}}$ number = 2x + 3
• $\sf {3^{rd}}$ number = 2x + 5

We know that the sum of the numbers is 147.

In simple words, we can say that, Given consecutive numbers = Total sum of numbers. Through this information we'll find out the value of x. And, at last, we'll find out numbers. Let's do it!

Finding value of x :-

${\sf \twoheadrightarrow \: \: \: \: \: {2x\ +\ 1\ +\ 2x\ +\ 3\ +\ 2x\ +\ 5\ =\ 147}}$

Adding the numbers and the like terms in the LHS.

$\twoheadrightarrow$ $\sf \: \: \: \: \: {6x\ +\ 9\ =\ 147}$

Putting 9 into RHS.

$\twoheadrightarrow$ $\sf \: \: \: \: \: {6x\ =\ 147\ -\ 9}$

Subtracting 9 from the given sum, i.e., 147.

$\twoheadrightarrow$ $\sf \: \: \: \: \: {6x\ =\ 138}$

Expressing RHS in fractional form. [The number 6 with variable x in LHS will be written in fractional form in RHS]

$\twoheadrightarrow$ $\sf \: \: \: \: \: {x\ =\ \dfrac{138}{6}}$

Dividing the fraction in RHS.

$\twoheadrightarrow$ $\: \: \: \: \: {\underline{\boxed{\purple{\textsf{\textbf{x\ =\ 23}}}}}}$

∴ Hence, the value of x is 23.

⠀⠀⠀⠀⠀━━━━━━━─━━━━━━━

Finding numbers :-

First NumbeR :-

$\twoheadrightarrow$ $\sf \: \: \: \: \: {2x\ +\ 1}$

Plugging in the value.

$\twoheadrightarrow$ $\sf \: \: \: \: \: {2(23)\ +\ 1}$

Multiplying the number in bracket with 2.

$\twoheadrightarrow$ $\sf \: \: \: \: \: {46\ +\ 1}$

Adding the numbers.

$\twoheadrightarrow$ $\: \: \: \: \: {\underline{\boxed{\red{\textsf{\textbf{First\ number\ =\ 47}}}}}}$

∴ Hence, first number is 47.

⠀⠀⠀⠀⠀━━━━━━━─━━━━━━━

Second NumbeR :-

$\twoheadrightarrow$ $\sf \: \: \: \: \: {2x\ +\ 3}$

Plugging in the value.

$\twoheadrightarrow$ $\sf \: \: \: \: \: {2(23)\ +\ 3}$

Multiplying the number in bracket with 2.

$\twoheadrightarrow$ $\sf \: \: \: \: \: {46\ +\ 3}$

Adding the numbers.

$\twoheadrightarrow$ $\: \: \: \: \: {\underline{\boxed{\red{\textsf{\textbf{Second\ number\ =\ 49}}}}}}$

∴ Hence, second number is 49.

⠀⠀⠀⠀⠀━━━━━━━─━━━━━━━

Third NumbeR :-

$\twoheadrightarrow$ $\sf \: \: \: \: \: {2x\ +\ 5}$

Plugging in the value.

$\twoheadrightarrow$ $\sf \: \: \: \: \: {2(23)\ +\ 5}$

Multiplying the number in bracket with 2.

$\twoheadrightarrow$ $\sf \: \: \: \: \: {46\ +\ 5}$

Adding the numbers.

$\twoheadrightarrow$ $\: \: \: \: \: {\underline{\boxed{\red{\textsf{\textbf{Third\ number\ =\ 51}}}}}}$

∴ Hence, third number is 51.