Question

three consecutive natural numbers are such that the sum of the square of the smallest number and the product of the other two, is 277. find the numbers​

Answer 1

$\huge{\underline{\purple{\rm{Solution:}}}}$

$\bold{\underline{\red{\rm{Let:}}}}$

• The 3 consecutive numbers are
• X,(X+1),(X+2

$\large{\underline{\green{\rm{To\:Find:}}}}$

• The number

$\large{\underline{\red{\rm{Given:}}}}$

• Three consecutive natural numbers are such that the sum of the square of the smallest number and the product of the other two, is 277.

$\small{\underline{\red{\rm{As\:per\:the\: above\: information:}}}}$

$\sf{\dashrightarrow X^2+(X+1)(X+2)=277}$

$\sf{\dashrightarrow X^2+X^2+3X+2=277}$

$\sf{\dashrightarrow 2X^2+3X+2=277}$

$\sf{\dashrightarrow 2X^2+3X-275=0}$

$\sf{\dashrightarrow 2X^2-22X+25X-275=0}$

$\sf{\dashrightarrow 2(X-11)+25(X-11)=0}$

$\sf{\dashrightarrow (2X+25)(X-11)=0}$

Therefore, X=-25/2 or 11

• here negative value can't be kept

Hence,

The number's are

$\sf{1st\: Number=X=11}$

$\sf{1st\: Number=X+1=11+1=12}$

$\sf{1st\: Number=X+2=11+2=13}$

Answer 2

Given:

Three consecutive natural numbers are such that the sum of the square of the smallest number and the product of the other two, is 277.

To Find:

All three numbers following above conditions

Given:

Let the three consecutive natural numbers be n,n+1 and n+2 where n is a natural number

So, according to the question

$\longrightarrow\mathrm{n^{2}+(n+1)(n+2)=277}$'

$\longrightarrow\mathrm{n^{2}+n^{2}+3n+2=277}$

$\longrightarrow\mathrm{2n^{2}+3n+2=277}$

$\longrightarrow\mathrm{2n^{2}+3n+2-277=0}$

$\longrightarrow\mathrm{2n^{2}+3n-275=0}$

$\longrightarrow\mathrm{2n^{2}+25n-22n-275=0}$

$\longrightarrow\mathrm{n(2n+25)-11(2n+25)=0}$

$\longrightarrow\mathrm{(n-11)(2n+25)=0}$

$\longrightarrow\mathrm{n=11,\dfrac{-25}{2}}$

Since, $\dfrac{-25}{2}$ is not a natural number

So, n=11

Therefore, the required numbers are

⇒ n=11

⇒ n+1=11+1=12

⇒ n+2=11+2=13

Hence, the required numbers are 11, 12 and 13.