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Find five consecutive odd integers if the sum of the first three integers is 3 more than the sum of the last two

Sagot :

Let the integers be:

  • 2k+1
  • 2k+3
  • 2k+5
  • 2k+7
  • 2k+9

It is one of the general forms to write an odd Integer..

ATQ,

[tex] \rm(2k + 1) + (2k + 3) + (2k + 5) = (2k + 7) + (2k + 9) + 3[/tex]

Using the equation Solve for K and then we will find the integers

[tex] \rm \: (2k + 1) + 2k + 3 + (2k + 5) = 2k + 7 + 2k + 9 + 3[/tex]

[tex] \rm2k + 1 + 2k + 2k + 5 = 2k + 7 + 2k + 9[/tex]

[tex] \rm \: 1 + 2k + 5 = 7 + 9[/tex]

[tex] \rm \: 6 + 2k = 16[/tex]

[tex] \rm \: 2k = 16 - 6[/tex]

[tex] \rm \: 2k = 10[/tex]

[tex] \boxed{ \tt \: k = 5}[/tex]

Now,

[tex]\large{|\underline{\mathtt{\red{1 ^{st} }\blue{ \: }\orange{i}\pink{n}\blue{t}\purple{e}\green{g}\red{e}\blue{r}\orange{ : }\green{ - }\red{}\purple{}\pink{}}}}[/tex]

[tex] \sf \: 2k + 1 \\ \sf 2 \times 5 + 1 \\ \sf \: 10 + 1 = 11[/tex]

[tex]\large{|\underline{\mathtt{\red{2 ^{nd} }\blue{ \: }\orange{i}\pink{n}\blue{t}\purple{e}\green{g}\red{e}\blue{r}\orange{ : }\green{ - }\red{}\purple{}\pink{}}}}[/tex]

[tex] \sf \: 2k + 3 \\ \sf 2 \times 5 + 3\\ \sf \: 10 + 3 = 13[/tex]

[tex]\large{|\underline{\mathtt{\red{3 ^{rd} }\blue{ \: }\orange{i}\pink{n}\blue{t}\purple{e}\green{g}\red{e}\blue{r}\orange{ : }\green{ - }\red{}\purple{}\pink{}}}}[/tex]

[tex] \sf \: 2k + 5 \\ \sf 2 \times 5 + 5\\ \sf \: 10 + 5 = 15[/tex]

[tex]\large{|\underline{\mathtt{\red{4^{th} }\blue{ \: }\orange{i}\pink{n}\blue{t}\purple{e}\green{g}\red{e}\blue{r}\orange{ : }\green{ - }\red{}\purple{}\pink{}}}}[/tex]

[tex] \sf \: 2k + 7 \\ \sf 2 \times 5 + 7\\ \sf \: 10 + 7= 17[/tex]

[tex]\large{|\underline{\mathtt{\red{5 ^{th} }\blue{ \: }\orange{i}\pink{n}\blue{t}\purple{e}\green{g}\red{e}\blue{r}\orange{ : }\green{ - }\red{}\purple{}\pink{}}}}[/tex]

[tex] \sf \: 2k + 9 \\ \sf 2 \times 5 + 9\\ \sf \: 10 + 9= 19[/tex]

➪ Thus, The five consecutive odd integers are 11,13,15,17,19...~

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