Is it true that u_n - u_(n-1) = d for an arithmetic sequence? Prove it and state when it fails.

Step-by-step solution

1. 1

   Recall the general term. An arithmetic sequence with first term $u_{1}$ and common difference $d$ has

   $u_{n}=u_{1}+(n-1)d\qquad (n\ge 1)$

   This is the closed form; the claim to be proved is the recursive form hiding inside it.

2. 2

   Write the previous term by replacing $n$ with $n-1$.

   $u_{n-1}=u_{1}+\left((n-1)-1\right)d=u_{1}+(n-2)d\qquad (n\ge 2)$

   The restriction $n\ge 2$ matters — there is no $u_{0}$ in a sequence indexed from $1$.

3. 3

   Subtract the two expressions. The $u_{1}$ terms cancel and the $d$-coefficients differ by exactly one:

   $u_{n}-u_{n-1}=\left[u_{1}+(n-1)d\right]-\left[u_{1}+(n-2)d\right]=\left[(n-1)-(n-2)\right]d=d$

   So yes, the identity holds for every $n\ge 2$, and the answer $d$ does not depend on $n$ — the gap is the same everywhere in the sequence.

4. 4

   Generalise to non-adjacent terms. The same subtraction with arbitrary indices gives

   $u_{m}-u_{n}=\left[u_{1}+(m-1)d\right]-\left[u_{1}+(n-1)d\right]=(m-n)d$

   with $u_{n}-u_{n-1}=d$ being the special case $m=n$, $n\to n-1$.

5. 5

   Check on a concrete sequence. With $u_{1}=3$ and $d=2$: $u_{2}=5$, $u_{3}=7$, $u_{4}=9$. Then $u_{4}-u_{3}=9-7=2=d$ ✓ and $u_{4}-u_{2}=9-5=4=(4-2)d$ ✓.

6. 6

   State the converse and the failure case. The identity is not just a consequence of being arithmetic — it characterises it: if $u_{n}-u_{n-1}$ equals the same constant for all $n$, the sequence is arithmetic. For a general sequence the difference need not be constant at all; e.g. for $u_{n}=n^{2}$ it is $u_{n}-u_{n-1}=2n-1$, which changes with $n$. So the formula may only be used once the sequence is known to be arithmetic.