Statement-1: `1+2+3....+n=(n(n+1))/(2),"for all "n in N`
Statement-2: `a+(a+d)+(a+2d)+....+(a+(n-1)d)=(n)/(2)[2a+(n-1)d]`

Answer

To solve the problem, we will prove both statements using the principle of mathematical induction.

### Step-by-Step Solution

#### Statement 1:
**Claim:** \(1 + 2 + 3 + \ldots + n = \frac{n(n+1)}{2}\) for all \(n \in \mathbb{N}\).

**Step 1: Base Case (n = 1)**
- For \(n = 1\):
- LHS: \(1\)
- RHS: \(\frac{1(1+1)}{2} = \frac{1 \cdot 2}{2} = 1\)
- Since LHS = RHS, the base case holds.

**Step 2: Inductive Hypothesis**
- Assume the statement is true for \(n = k\):
\[ 1 + 2 + 3 + \ldots + k = \frac{k(k+1)}{2} \]

**Step 3: Inductive Step (Prove for n = k + 1)**
- We need to show that:
\[ 1 + 2 + 3 + \ldots + k + (k + 1) = \frac{(k + 1)(k + 2)}{2} \]
- Starting from the inductive hypothesis:
\[ 1 + 2 + 3 + \ldots + k + (k + 1) = \frac{k(k + 1)}{2} + (k + 1) \]
- Combine the terms on the right:
\[ = \frac{k(k + 1)}{2} + \frac{2(k + 1)}{2} = \frac{k(k + 1) + 2(k + 1)}{2} = \frac{(k + 1)(k + 2)}{2} \]
- Thus, the statement holds for \(n = k + 1\).

**Conclusion for Statement 1:**
By the principle of mathematical induction, the statement is true for all \(n \in \mathbb{N}\).

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#### Statement 2:
**Claim:** \(a + (a + d) + (a + 2d) + \ldots + (a + (n-1)d) = \frac{n}{2}[2a + (n-1)d]\).

**Step 1: Base Case (n = 1)**
- For \(n = 1\):
- LHS: \(a\)
- RHS: \(\frac{1}{2}[2a + (1-1)d] = \frac{1}{2}[2a + 0] = a\)
- Since LHS = RHS, the base case holds.

**Step 2: Inductive Hypothesis**
- Assume the statement is true for \(n = k\):
\[ a + (a + d) + (a + 2d) + \ldots + (a + (k-1)d) = \frac{k}{2}[2a + (k-1)d] \]

**Step 3: Inductive Step (Prove for n = k + 1)**
- We need to show that:
\[ a + (a + d) + (a + 2d) + \ldots + (a + (k-1)d) + (a + kd) = \frac{k + 1}{2}[2a + kd] \]
- Starting from the inductive hypothesis:
\[ a + (a + d) + \ldots + (a + (k-1)d) + (a + kd) = \frac{k}{2}[2a + (k-1)d] + (a + kd) \]
- Combine the terms:
\[ = \frac{k}{2}[2a + (k-1)d] + \frac{2(a + kd)}{2} = \frac{k[2a + (k-1)d] + 2a + 2kd}{2} \]
- Simplifying the right side:
\[ = \frac{(k + 1)[2a + kd]}{2} \]
- Thus, the statement holds for \(n = k + 1\).

**Conclusion for Statement 2:**
By the principle of mathematical induction, the statement is true for all \(n \in \mathbb{N}\).

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