Question

The sum of three consecutive multiples of 11 is 363. Find the multiples

Answer 1

Answer:

multiples are 110,121,132

Step-by-step explanation:

Let no. be x, x+11, x+22

a.t.q

x+x+11+x+22=363

=3x+33=363

=3x = 363-33

3x = 330

x=110

so multiples are 110,121,132

Answer 2

Answer:-

$\sf{The \ multiples \ of \ 11 \ are \ 110, \ 121}$

$\sf{ and \ 132 \ respectively. }$

Given:

• The sum of three consecutive multiples of 11 is 363.

To find:

• The multiples.

Solution:-

$\sf{Let \ the \ constant \ be \ n.}$

$\sf{\therefore{Three \ consecutive \ multiples \ of \ 11 \ are}}$

$\sf{11(n-1), \ 11n \ and \ 11(n+1)}$

$\sf{According \ to \ the \ given \ condition. }$

$\sf{11(n-1)+11n+11(n+1)=363}$

$\sf{\therefore{11(n-1+n+n+1)=363}}$

$\sf{\therefore{3n=\frac{363}{11}}}$

$\sf{\therefore{3n=33}}$

$\sf{\therefore{n=\frac{33}{3}}}$

$\boxed{\sf{\therefore{n=11}}}$

$\sf{The \ multiples \ of \ 11 \ are}$

$\sf{\implies{11(n-1)=11(11-1)=11(10)=110,}}$

$\sf{\implies{11(n)=11(11)=121,}}$

$\sf{\implies{11(n+1)=11(11+1)=11(12)=132.}}$

$\sf\purple{\tt{\therefore{The \ multiples \ of \ 11 \ are \ 110, \ 121}}}$

$\sf\purple{\tt{and \ 132 \ respectively. }}$