The sum of the first k terms of an arithmetic progression (AP) is given by the formula S_k = (k)/(2)[2a + (k-1)d], where a is the first term and d is the common difference.

If the sum of the first $m$ terms of an AP is the same as the sum of its first $n$ terms ( $m \neq n$ ), show that the sum of its first $(m + n)$ terms is: $S_{m+n} = 0$

Answer: $S_{m+n} = 0$

Step-by-step solution

Step 1: Write the formula for the sum of an AP

The sum of the first $k$ terms of an arithmetic progression (AP) is given by the formula $S_k = \frac{k}{2}[2a + (k-1)d]$ , where $a$ is the first term and $d$ is the common difference.

Step 2: Set up the given condition

We are given that the sum of the first $m$ terms is equal to the sum of the first $n$ terms. We substitute $k=m$ and $k=n$ into the sum formula and set them equal.

Step 3: Simplify the equation

First, multiply both sides by 2 to clear the denominators. Then, rearrange the terms to group $2a$ and $d$ . Factor out common terms, noting that $m^2 - n^2 = (m-n)(m+n)$ . Since $m \neq n$ , we know that $m-n \neq 0$ , allowing us to divide by $(m-n)$ .

Step 4: Derive the relationship between 'a' and 'd'

Since $m \neq n$ , we can divide both sides of the equation by $(m-n)$ . This gives us a crucial relationship between the first term $a$ , the common difference $d$ , and the numbers of terms $m$ and $n$ .

Step 5: Calculate the sum of the first (m+n) terms

Now, we need to find the sum of the first $(m+n)$ terms. We use the sum formula again, substituting $k = (m+n)$ .

Step 6: Substitute the derived relationship

From the previous step, we found that $2a + (m+n-1)d = 0$ . We substitute this into the expression for $S_{m+n}$ . Multiplying by zero results in zero.

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If the sum of the first mmm terms of an AP is the same as the sum of its first nnn terms (m≠nm \neq nm=n), show that the sum of its first (m+n)(m + n)(m+n) terms is: Sm+n=0S_{m+n} = 0Sm+n​=0