Question

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

Answer 1

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,

$\rm(2k + 1) + (2k + 3) + (2k + 5) = (2k + 7) + (2k + 9) + 3$

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

$\rm \: (2k + 1) + 2k + 3 + (2k + 5) = 2k + 7 + 2k + 9 + 3$

$\rm2k + 1 + 2k + 2k + 5 = 2k + 7 + 2k + 9$

$\rm \: 1 + 2k + 5 = 7 + 9$

$\rm \: 6 + 2k = 16$

$\rm \: 2k = 16 - 6$

$\rm \: 2k = 10$

$\boxed{ \tt \: k = 5}$

Now,

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

$\sf \: 2k + 1 \\ \sf 2 \times 5 + 1 \\ \sf \: 10 + 1 = 11$

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

$\sf \: 2k + 3 \\ \sf 2 \times 5 + 3\\ \sf \: 10 + 3 = 13$

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

$\sf \: 2k + 5 \\ \sf 2 \times 5 + 5\\ \sf \: 10 + 5 = 15$

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

$\sf \: 2k + 7 \\ \sf 2 \times 5 + 7\\ \sf \: 10 + 7= 17$

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

$\sf \: 2k + 9 \\ \sf 2 \times 5 + 9\\ \sf \: 10 + 9= 19$

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