Find sum of first (m-n) terms of AP given swapped conditions

[subhradeep mahata]

Homework Statement

In an AP, sum of first n terms is equal to m and sum of first m terms is equal to n. Then, find the sum of first (m-n) terms in terms of m and n, assuming m>n.

Homework Equations

Sum of an AP: n/2 * {2a+ (n-1)d}

The Attempt at a Solution

We get two equations:
m= n/2 * {2a+ (n-1)d} and
n= m/2 * {2a+ (m-1)d}
where a and d are first term and common difference respectively. Now, we can solve for a and d separately and plug the values in this equation:
S_m-n = (m-n)/2 * {2a + (m-n-1)d}
But, as we can see it is a very lengthy and ambiguous method. Can someone please find an easier method for the problem?

 

[Ray Vickson]

Homework Statement

In an AP, sum of first n terms is equal to m and sum of first m terms is equal to n. Then, find the sum of first (m-n) terms in terms of m and n, assuming m>n.

Homework Equations

Sum of an AP: n/2 * {2a+ (n-1)d}

The Attempt at a Solution

We get two equations:
m= n/2 * {2a+ (n-1)d} and
n= m/2 * {2a+ (m-1)d}
where a and d are first term and common difference respectively. Now, we can solve for a and d separately and plug the values in this equation:
S_m-n = (m-n)/2 * {2a + (m-n-1)d}
But, as we can see it is a very lengthy and ambiguous method. Can someone please find an easier method for the problem?

Certainly it is lengthy, but there is nothing "ambiguous" about it. Anyway, what is the final answer that you obtained?

 

[subhradeep mahata]

I got this after a tedious calculation : {(m-n)(m+2n)}/m
please let me know if there are any easier methods.

 

[RPinPA]

It looks like subtracting those two equations gives you something very close to what you want.
$$m = \frac{n}{2} \left [2a + (n-1)d \right ] = na + \frac {n(n-1)d}{2} \\
n = \frac{m}{2} \left [2a + (m-1)d \right ] = ma + \frac {m(m-1)d}{2} \\
m - n = (n - m)a + \left [n(n-1) - m(m-1) \right ] \frac {d}{2}$$
$$S_{m-n} = \frac {m-n}{2} \left [2a + (m - n - 1)d \right ] = (m - n)a + (m-n)(m - n - 1)\frac {d}{2}$$

So it's not hard to express $S_{m-n}$ in terms of $(m - n)$ and $d$. That may be useful in shortening the calculation.

 

[verty]

Here is another shortening:
$$S_{m-n} = {m - n \over 2} (2a + (m - n - 1)d) \\
= {m \over 2} (2a + (m-1)d - nd) - {n \over 2} (2a + (n-1)d - (2n - m)d) \\
= (n - m) + {d \over 2} (n(2n - m) - mn) \\
= (n - m) + {d \over 2} 2n (n - m) \\
= (n - m) (1 + nd)$$