(A)OR(B)

Find the sum of integers between $\displaystyle 100$ and $\displaystyle 200$ which are (i) divisible by $\displaystyle 9$ (ii) not divisible by $\displaystyle 9$ .

Integers divisible by $\displaystyle 9$ are $\displaystyle 108$, $\displaystyle 117$, 126...., $\displaystyle 198$

\[\begin{aligned} & a=108, d=9 \\ & a+(n-1) d=198 \\ & \Rightarrow 108+(n-1) 9=198 \Rightarrow n=11 \\ & \begin{aligned} \mathrm{S}_{11}=\frac{n}{2}(\mathrm{a}+l) & =\frac{11}{2}(108+198) \\ & =1683 \end{aligned} \\ & \text { (ii) Integers are } 101,102,103, \ldots \ldots, 199 \end{aligned} \]

Sum of all integers \(\displaystyle =\frac{99}{2}(101+199)\)

\[=\frac{99}{2} \times 300=14850 \]

Sum of integers not divisible by \(\displaystyle 9=14850-1683\)

\[\text { = } 13167 \]

\[\begin{aligned} & -4+(-1)+2+5+\ldots \ldots+x=437 \\ & \text { Here } a=-4, d=3 \\ & -4+(n-1) 3=x \Rightarrow n=\frac{x+7}{3} \\ & S_{n}=437 \\ & \Rightarrow\left(\frac{x+7}{3}\right) \cdot \frac{1}{2}(-4+x)=437 \\ & x^{2}+3 x-28=437 \times 6=2622 \\ & x^{2}+3 x-2650=0 \\ & (x+53)(x-50)=0 \\ & x \neq-53, x=50 \end{aligned} \]