(a) If the sum of first m terms of an A.P. is same as sum of its first n terms ( m ≠ n ), then show that the sum of its first ( m + n ) terms is zero.

\[\begin{aligned} & S_{m}=S_{n} \\ & \Rightarrow \frac{m}{2}\left[2 a+(m-1) d=\frac{n}{2}[2 a+(n-1) d]\right. \\ & \Rightarrow 2 a(m-n)=d\left(n^{2}-m^{2}\right)-d(n-m) \\ & \Rightarrow 2 a=-d(m+n-1) \\ & \text { or } 2 a+(m+n-1) d=0 \\ & \text { i.e., } S_{m+n}=\frac{m+n}{2}[2 a+(m+n-1) d]=0 \end{aligned} \]

Let the numbers be a - d, a, a + d

\[\begin{aligned} & \therefore a-d+a+a+d=24 \\ & \Rightarrow a=8 \end{aligned} \]

Also, \(\displaystyle (\mathrm{a}-\mathrm{d})^{2}+\mathrm{a}^{2}+(\mathrm{a}+\mathrm{d})^{2}=194\)

\[\begin{aligned} & \Rightarrow(8-d)^{2}+8^{2}+(8+d)^{2}=194 \\ & \Rightarrow d^{2}=1 \Rightarrow d= \pm 1 \\ & \therefore \text { Numbers are } 7,8,9 \text { or } 9,8,7 \end{aligned} \]