Question

find three consequtive integers such that third is one and a half times the first

Answer 1

$\underline{ \underline{ \Large \pmb{\mathit{ {Required \: Solution:}} }} }$

✰ Here, we have to find three consequtive integers , where third is one and a half times the first. Firstly let's understand the meaning of the consecutive integers. "Consecutive" means following something in sequence and "integers" refers to the numbers including negative numbers, 0 and positive numbers. So, consecutive integers means tthe integers follow each other in sequence. In order to solve this problem, we'll first assume the numbers as variables and then we'll form an algebraic equation and by solving that equation we'll find those three consecutive integers.

⠀⠀⠀⠀⠀_____________

Let,

• First integer = x

• Second integer = (x + 1)

• Third integer = (x + 2)

As per the given question,

$\longrightarrow \sf { Third \: Integer = 1 \dfrac{1}{2} \times First \: Integer }$

Substituting the values,

$\longrightarrow \sf { x +2 = 1 \dfrac{1}{2}x }$

$\longrightarrow \sf { x +2 = \dfrac{3}{2}x }$

$\longrightarrow \sf { x +2 = \dfrac{3x}{2} }$

Transposing 2 from R.H.S to L.H.S.

$\longrightarrow \sf { 2(x +2) = 3x}$

$\longrightarrow \sf { 2x + 4 = 3x}$

$\longrightarrow \sf { 4 = 3x - 2x}$

$\longrightarrow \sf { 4 = x}$ . . . ★

Therefore, the three consecutive integers are as follows :-

$\longrightarrow \sf { First \: Integer = x}$

$\longrightarrow \boxed{\pmb{\rm \red{ First \: Integer = 4}}}$

$\longrightarrow \sf { Second \: Integer = (x + 1)}$

$\longrightarrow \sf { Second \: Integer = (4 + 1)}$

$\longrightarrow \boxed{\pmb{\rm \red{ Second \: Integer = 5}}}$

$\longrightarrow \sf { Third \: Integer = (x + 2)}$

$\longrightarrow \sf { Third \: Integer = (4 + 2)}$

$\longrightarrow \boxed{\pmb{\rm \red{ Third \: Integer = 6}}}$

Therefore, the three consecutive integers are 4,5,6 such that third is one and a half times the first.

Answer 2

$\mathfrak{\large{\pmb{Given}}}\begin{cases} {\sf three\:consecutive\:integers} \\ { \sf third\: integer\:is\:one\:and\:half\:the\:first} \end{cases}$

${\large{\mathfrak{\pmb{To\:find :}}}} \:\: \sf{the\:three\:consecutive\:integers}$

$\large\mathfrak{\pmb{Solution:}}$

• Let the three consecutive integers be $\sf{{\pmb{x}}~,~~{\pmb{x+1}}~~and~~{\pmb{x+2}}}$
• As per the question, third integer is one and a half times the first

$\sf\implies{x+2=1{\dfrac{1}{2}}×~x}$

$\sf\implies{x+2={\dfrac{3}{2}}×~x}$

$\sf\implies{x+2={\dfrac{3x}{2}}}$

$\sf\implies{2(x+2)=3x}$

$\sf\implies{2x+4=3x}$

$\sf\implies{4=3x-2x}$

$\sf\implies{3x-2x=4}$

$\sf{~~~~~ {\purple{ \bigstar~ {\underline{\boxed{\sf{\pmb{x=4}}}}}}}}$

$\large{\therefore {\mathfrak{\pmb{Required~answer:}}}}$

$\purple{ \bigstar~ {\sf{\pmb{First~integer:}}}}$ $\sf{ x = {\underline{\boxed{\sf{\pmb{~4~}}}}}}$

$\purple{ \bigstar ~{\sf{\pmb{Second~integer:}}}}$ $\sf{ x+1 =4+1= {\underline{\boxed{\sf{\pmb{~5~}}}}}}$

$\purple{ \bigstar~ {\sf{\pmb{Third~integer:}}}}$ $\sf{ x+2 = 4+2={\underline{\boxed{\sf{\pmb{~6~}}}}}}$