Question

b) Find the sum of all two-digit numbers which are multiple of 11.​

Answer 1

sum=484

.............

Answer 2

Answer:

$\sf{The \ sum \ of \ all \ two \ digits \ number \ which}$

$\sf{are \ divisible \ by \ 11 \ is \ 495.}$

To find:

$\sf{Sum \ of \ all \ two \ digits \ number \ which \ are}$

$\sf{multiple \ of \ 11.}$

Solution:

$\sf{Two \ digit \ numbers \ which \ are \ multiple \ of \ 11 \ are}$

$\sf{11, \ 22, \ 33,..., \ 99}$

$\sf{It \ forms \ an \ A.P.}$

$\sf{Here,}$

$\sf{a=11, \ d=11 \ and \ t_{n}=99}$

$\boxed{\sf{t_{n}=a+(n-1)d}}$

$\sf{\therefore{99=11+(n-1)\times11}}$

$\sf{\therefore{11(n-1)=88}}$

$\sf{\therefore{n-1=\dfrac{99}{11}}}$

$\sf{\therefore{n=8+1}}$

$\sf{\therefore{n=9}}$

$\sf{By \ the \ formula \ of \ n \ terms \ of \ an \ A.P.}$

$\boxed{\sf{S_{n}=\dfrac{n}{2}[t_{1}+t_{n}}}$

$\sf{\therefore{S_{9}=\dfrac{9}{2}[11+99]}}$

$\sf{\therefore{S_{9}=\dfrac{9}{2}\times110}}$

$\sf{\therefore{S_{9}=9\times55}}$

$\sf{\therefore{S_{9}=495}}$

$\sf\purple{\tt{\therefore{The \ sum \ of \ all \ two \ digits \ number \ which}}}$

$\sf\purple{\tt{are \ divisible \ by \ 11 \ is \ 495.}}$