Question

The sum of three consecutive multiples of 7 is 84 . What are the numbers ??​

Answer 1

Answer:

hope this will help you

Answer 2

Answer:

Given :-

• The sum of three consecutive multiples of 7 is 84.

To Find :-

• What are the numbers.

Solution :-

Let,

$\mapsto \sf\bold{First\: consecutive\: integers =\: 7x}$

$\mapsto \sf\bold{Second\: consecutive\: integers =\: 7(x + 1)}$

$\mapsto \sf\bold{Third\: consecutive\: integers =\: 7(x + 2)}$

According to the question,

$\implies \sf 7x + 7(x + 1) + 7(x + 2) =\: 84$

$\implies \sf 7x + 7x + 7 + 7x + 14 =\: 84$

$\implies \sf 7x + 7x + 7x + 7 + 14 =\: 84$

$\implies \sf 21x + 21 =\: 84$

$\implies \sf 21x =\: 84 - 21$

$\implies \sf 21x =\: 63$

$\implies \sf x =\: \dfrac{\cancel{63}}{\cancel{21}}$

$\implies \sf\bold{\purple{x =\: 3}}$

Hence, the required numbers are :

$\mapsto$ First consecutive integers :

$\longrightarrow \sf 7x$

$\longrightarrow \sf 7(3)$

$\longrightarrow \sf\bold{\red{21}}$

$\mapsto$ Second consecutive integers :

$\longrightarrow \sf 7(x + 1)$

$\longrightarrow \sf 7(3 + 1)$

$\longrightarrow \sf 21 + 7$

$\longrightarrow \sf\bold{\red{28}}$

$\mapsto$ Third consecutive integers :

$\longrightarrow \sf 7(x + 2)$

$\longrightarrow \sf 7(3 + 2)$

$\longrightarrow \sf 21 + 14$

$\longrightarrow \sf\bold{\red{35}}$

$\therefore$ The numbers are 21, 28 and 35 respectively.

VERIFICATION :-

$\leadsto \sf 7x + 7(x + 1) + 7(x + 2) =\: 84$

By putting x = 3 we get,

$\leadsto \sf 7(3) + 7(3 + 1) + 7(3 + 2) =\: 84$

$\leadsto \sf 21 + 7(4) + 7(5) =\: 84$

$\leadsto \sf 21 + 28 + 35 =\: 84$

$\leadsto \sf\bold{\pink{84 =\: 84}}$

Hence, Verified.