Porisma

Worked solution

Relate three arithmetic terms and test a proof

(a) Suppose x,y,zx,y,z are three consecutive terms of an AP with common difference dd. Write an equation relating x,y,zx,y,z. (b) Consider: “Since un=2+3nu_n=2+3n, we have u1=5,u2=8,u3=11u_1=5,u_2=8,u_3=11. Then u2−u1=u3−u2=3u_2-u_1=u_3-u_2=3, a constant, so the sequence is an AP.” Is this argument valid? Why?

CJC · Sequences and Series · Unnumbered AP questions, page 13

Three consecutive AP terms

Answer

y=(x+z)/2y=(x+z)/2.

Equate consecutive differences

y−x=z−y=d⟹2y=x+z⟹y=x+z2y-x=z-y=d\quad\Longrightarrow\quad2y=x+z\quad\Longrightarrow\quad y=\frac{x+z}{2}

The middle term is the arithmetic mean of its neighbours. The source uses this relation to solve AP problems.

How did this part go?

Test the proposed argument

Answer

The argument is invalid: it checks only two pairs, rather than every pair of consecutive terms.

Distinguish the conclusion from its justification

The printed explanation requires the general term and its neighbour, so varying n covers all valid pairs. The sequence really is an AP, but those first three terms alone do not prove it.

Complete a proof for every positive index

un+1−un=(2+3(n+1))−(2+3n)=3(n∈Z+)u_{n+1}-u_n=(2+3(n+1))-(2+3n)=3\qquad(n\in\mathbb Z^+)

This general calculation is authored working that completes the source explanation.

How did this part go?

Common mistakes

An infinite sequence is not proved arithmetic by checking a finite prefix.