(a) In an A.P., the first term is 4 and the last term is 31. If sum of all the terms is 175, find the number of terms and the common difference.

Here \(\displaystyle \mathrm{a}=4, l=31\)

\[\begin{aligned} & S_{n}=175 \\ \Rightarrow & \frac{n}{2}(4+31)=175 \\ \Rightarrow & n=10 \end{aligned} \]

So, \(\displaystyle \mathrm{a}_{10}=31\)

\[\begin{aligned} & \Rightarrow 4+9 d=31 \\ & \Rightarrow d=3 \end{aligned} \]

\[\begin{aligned} d & =18-21=-3 \\ & S_{n}=0 \\ \Rightarrow & \frac{n}{2}[2 \times 21+(n-1) \times(-3)]=0 \\ \Rightarrow & n^{2}-15 n=0 \Rightarrow n(n-15)=0 \\ \Rightarrow & n=0,15 \end{aligned} \]

Since 'n' is a natural number.

\[\therefore \mathrm{n}=15 \]