A standard arithmetic-progression identity problem, not tied to a specific exam paper.
Problem
In an AP, if $m\cdot t_m=n\cdot t_n$ (with $m\neq n$), find $t_{m+n}$.
Key insight. Writing both terms using the standard $t_k=a+(k-1)d$ form and expanding the given condition, every term collects around a common factor of $(m-n)$ — and since $m\neq n$ is given, that factor can be safely divided out, collapsing the whole equation down to exactly the expression for $t_{m+n}$.
Watch this explained step by step → · More on the Shiwam’s Classes channel
Approach
Write $t_m=a+(m-1)d$ and $t_n=a+(n-1)d$, and expand the given condition $m\cdot t_m=n\cdot t_n$. Collect all terms on one side and factor — the expression will factor neatly into $(m-n)$ times something. Since $m\neq n$, divide out $(m-n)$, and recognise what remains as exactly $t_{m+n}$.
Solution
Step 1 — Write the condition using tₖ = a + (k−1)d
$$m\left[a+(m-1)d\right] = n\left[a+(n-1)d\right]$$
Step 2 — Expand both sides
$$ma+m(m-1)d = na+n(n-1)d$$
Step 3 — Move everything to one side
$$a(m-n) + \left[m(m-1)-n(n-1)\right]d = 0$$
Step 4 — Simplify the bracketed coefficient of d
$$m(m-1)-n(n-1) = m^2-m-n^2+n = (m^2-n^2)-(m-n) = (m-n)(m+n)-(m-n)$$
$$= (m-n)\left[(m+n)-1\right]$$
Step 5 — Substitute back
$$a(m-n) + (m-n)(m+n-1)d = 0$$
$$(m-n)\left[a+(m+n-1)d\right] = 0$$
Step 6 — Divide out (m − n), since m ≠ n
$$a+(m+n-1)d = 0$$
Step 7 — Recognise this as tₘ₊ₙ
By definition, $t_{m+n}=a+\left[(m+n)-1\right]d$ — exactly the left-hand side of Step 6:
$$t_{m+n} = 0$$
Answer
$$t_{m+n} = 0$$
Common mistakes
- Dividing by $(m-n)$ without first confirming $m\neq n$. The problem’s own condition $m\neq n$ is what licenses this step — without it, $(m-n)$ could be zero, and the factored equation $(m-n)[\ldots]=0$ would be automatically true regardless of the bracketed expression, giving no information about $t_{m+n}$ at all.
- Losing track of which terms combine into $(m-n)(m+n-1)$. The algebra in Step 4 involves factoring $m^2-n^2$ as a difference of squares and separately extracting a second copy of $(m-n)$ from the $-m+n$ part — missing either factoring step leaves the expression looking irreducibly messy.
Practise next
- In a GP, if $t_m^n = t_n^m$ for $m\neq n$, express what this implies about the common ratio $r$, using a similar factor-and-cancel approach.
Show answer
The first term equals the common ratio, so $t_k=r^k$. With $t_k=ar^{k-1}$, the condition $t_m^{\,n}=t_n^{\,m}$ reads $a^{n}r^{n(m-1)}=a^{m}r^{m(n-1)}$.
Collect powers: $a^{n-m}=r^{\,m(n-1)-n(m-1)}=r^{\,n-m}$, so $\left(\dfrac ar\right)^{n-m}=1$.
Since $m\neq n$, and taking positive terms, this forces $a=r$. Every term is then $t_k=r\cdot r^{k-1}=r^{k}$ — the GP counterpart of the AP result above, where the condition forced $t_{m+n}=0$.
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.