Question

Find 3 consecutive positive even integers whose sum is 90.​

Answer 1

Answer:

a = 29

Step-by-step explanation:

a + a+ 1 + a + 2 = 90

3a + 3 =90

3[a+ 1] =90

a + 1 = 90/3

a + 1 = 30

a =29

first no = 29

second no = 30

third no = 31

Answer 2

$\Large{\textbf{\underline{\underline{According\:to\:the\:Question}}}}$

Assumption

The numbers be

p , p + 2 , p + 4

According to the information,

Sum of the numbers = 90

Situation would be like :-

p + p + 2 + p + 4 = 90

3p + 6 = 90

3p = 90 - 6

3p = 84

${\boxed{\sf\:{p=\dfrac{84}{3}}}}$

p = 28

Therefore we get,

p = 28

p + 2 = 28 + 2 = 30

p + 4 = 28 + 4 = 32

Hence,

Numbers are 28, 30 and 32

$\boxed{\begin{minipage}{13 cm} Additional Information \\ \\ \ A\; Quadratic\; Equation\;has\;three\;equal\;roots \\ \\ 1)Real\;and\;Distinct \\ \\ 2)Real\;and\;Coincident \\ \\ 3) Imaginary \\ \\ If\;p(x)\;is\;a\; quadratic\; polynomial\;then\;p(x)=0\;is\;called\; Quadratic\; Polynomial \\ \\ General\;Formula=ax^2+bx+c=0 \\ \\ A\;polynomial\;whose\;degree\;will\;be\;2\;is\; considered\;as\; Quadratic\; Polynomial \\ \\ Rules\;for\; solving\; Quadratic\; Equations:- \\ \\ Put\;all\;the\;terms\;into\;RHS\;and\;make\it\;zero \\ \\ Substitute\;all\; factors\;equal\;to\;Zero\;Get\;a\;equal\; solution \end{minipage}}$