a + (a + d) + (a + 2d) + ... (a + (n − 1) d)

Solution:

Let P(n) be the given statement.

Now,

$P(n): a+(a+d)+(a+2 d)+\ldots+(a+(n-1) d)=\frac{n}{2}[2 a+(n-1) d]$

Step 1:

$P(1)=a=\frac{1}{2}(2 a+(1-1) d)$

Hence, $P(1)$ is true.

Step 2:

Suppose $P(m)$ is true.

Then,

$a+(a+d)+\ldots+(a+(m-1) d)=\frac{m}{2}[2 a+(m-1) d]$

We have to show that $P(m+1)$ is true whenever $P(m)$ is true.

That is,

$a+(a+d)+\ldots+(a+m d)=\frac{(m+1)}{2}[2 a+m d]$

$W e$ know that $P(m)$ is true.

Thus, we have :

$a+(a+d)+\ldots+(a+(m-1) d)=\frac{m}{2}[2 a+(m-1) d]$

$\Rightarrow a+(a+d)+\ldots+(a+(m-1) d)+(a+m d)=\frac{m}{2}[2 a+(m-1) d]+(a+m d)$

[Adding $(a+m d)$ to both sides]

$\Rightarrow P(m+1)=\frac{1}{2}\left[2 a m+m^{2} d-m d+2 a+2 m d\right]$

$\Rightarrow P(m+1)=\frac{1}{2}[2 a(m+1)+m d(m+1)]$

$=\frac{1}{2}(m+1)(2 a+m d)$

Thus, $P(m+1)$ is true.

By the principle of $m$ athematical $i$ nduction, $P(n)$ is true for all $n \in N$.