A general problem-solving technique video — not a specific exam question.
Problem
If $S_m=S_n$ for some AP (where $m\neq n$), prove that $S_{m+n}=0$.
Key insight. Writing both $S_m$ and $S_n$ using the standard sum formula and setting them equal produces one equation relating $2a$ and $d$. That same relationship, substituted back into the formula for $S_{m+n}$, makes the bracket inside collapse to exactly zero — the proof doesn’t need $m$ and $n$ individually, only their combination.
Watch this explained step by step → · More on the Shiwam’s Classes channel
Approach
Write $S_m$ and $S_n$ using the standard AP sum formula, set them equal (as given), and simplify to find a relationship between $2a$ and $d$. Then substitute this relationship into the formula for $S_{m+n}$ and show the bracket vanishes, forcing $S_{m+n}=0$.
Solution
Step 1 — Write Sm = Sn using the sum formula
$$\frac{m}{2}\big[2a+(m-1)d\big] = \frac{n}{2}\big[2a+(n-1)d\big]$$
Cancelling the $\frac{1}{2}$:
$$m\big[2a+(m-1)d\big] = n\big[2a+(n-1)d\big]$$
Step 2 — Expand both sides
$$2am+m(m-1)d = 2an+n(n-1)d$$
$$2a(m-n) = d\big[n(n-1)-m(m-1)\big] = d(n^2-n-m^2+m)$$
Step 3 — Factor the right-hand side
$$n^2-m^2-n+m = -(m^2-n^2)+(m-n) = -(m-n)(m+n)+(m-n) = (m-n)\big[1-(m+n)\big]$$
So:
$$2a(m-n) = d(m-n)\big[1-(m+n)\big]$$
Step 4 — Divide by (m − n)
Since $m\neq n$, dividing both sides by $(m-n)$:
$$2a = d\big[1-(m+n)\big]$$
Step 5 — Write Sm+n using the sum formula
$$S_{m+n} = \frac{m+n}{2}\big[2a+(m+n-1)d\big]$$
Step 6 — Substitute the relationship from Step 4
$$2a+(m+n-1)d = d\big[1-(m+n)\big] + (m+n-1)d$$
$$= d – d(m+n) + (m+n)d – d = 0$$
Step 7 — Conclude
Since the bracket is zero:
$$S_{m+n} = \frac{m+n}{2}\times0 = 0$$
Answer
$$S_{m+n} = 0$$
Common mistakes
- Trying to solve for $a$ and $d$ individually. The proof only ever needs the combination $2a$ in terms of $d$ — chasing separate numeric values for $a$ and $d$ is unnecessary and often impossible from the given information alone.
- Dividing by $(m-n)$ without first confirming $m\neq n$. This is given in the problem, but it’s worth explicitly stating why the division is valid — if $m=n$, the original equation $S_m=S_n$ would be trivially true and give no information.
Practise next
- If $S_p = S_q$ for some AP with $p \neq q$, and additionally $S_p = k$ for a known constant $k$, find $a$ and $d$ in terms of $p$, $q$, and $k$, using the same relationship derived here.
Show answer
$d=-\dfrac{2k}{pq}$ and $a=\dfrac{k(p+q-1)}{pq}$.
$S_p=S_q$ with $p\neq q$ forces $2a+(p+q-1)d=0$ — that is the relation behind $S_{p+q}=0$.
The extra condition $S_p=\dfrac{p}{2}\left[2a+(p-1)d\right]=k$ gives $2a+(p-1)d=\dfrac{2k}{p}$. Subtracting the first relation leaves $-qd=\dfrac{2k}{p}$, so $d=-\dfrac{2k}{pq}$.
Substituting back, $2a=-(p+q-1)d=\dfrac{2k(p+q-1)}{pq}$, giving $a$ as above. Two conditions, two unknowns — the vanishing $S_{p+q}$ is what makes the first one usable.
Doubts are answered by Shiwam, usually within a day. Ask about this question specifically — a general question about the chapter is better asked in class.