Question

Find three consecutive whose some is 81.

Answer 1

Answer:

26,27,28

Step-by-step explanation:

let the 1st no be X

let the second no be X+1

let the third no be X+2

then a/q

x+(x+1) +(x+2) = 81

3x +3 = 81

3x =78

x=26

so 1 no = 26

2 no =26+1=

27

and 28

Answer 2

Hey friend,

Here is the solution :-

$let \: the \: first \: number \: be : x \\ then \: second \: number \: be : x + 1 \\ third \: number \: be : \: \: \: \: \: \: \: \: \: \: \: \: x + 2 \\ \\ their \: sum \: is \: 81 \: so : - \\ \\ x + (x + 1) + (x + 2) = 81 \\ \\ x + x + 1 + x + 2 = 81 \\ \\ 3x + 3 = 81 \\ \\ 3x = 81 - 3 \\ \\ 3x = 78 \\ \\ x = \frac{78}{3} \\ \\ x = 26$

$now \: first \: number = x = 26 \\ \\ second \: number \: = x + 1 = 26 + 1 = 27 \\ \\ third \: number \: = x + 2 = 26 + 2 = 28$

So, the numbers are 26, 27 & 28.

$verification : - \\ \\ according \: to \: question \: the \: sum \: of \: those \: \\ numbers \: should \: be \: 81 : \\ \\ 26 + 27 + 28 = 81 \\ \\ 81 = 81 \: \: \: \: \: \: \: (verified)$

✨I hope this will help you....✨