Exercises — Arithmetic progression (AP) — nth term, sum of n terms — derivations

The test: a sequence is an AP only if every consecutive difference is the same number. Subtract neighbour from neighbour and check they all match.

(a) $12-7=5$, $17-12=5$, $22-17=5$. All equal ⇒ AP with $a=7$, $d=5$. (b) $4-2=2$ but $8-4=4$. Differences change ⇒ not an AP (this one multiplies by 2 — that's a Geometric Progression (GP)). (c) $7-10=-3$, $4-7=-3$, $1-4=-3$. All equal ⇒ AP with $a=10$, $d=-3$. Note $d$ is allowed to be negative — the sequence goes downhill. (d) $5-5=0$ every time ⇒ AP with $a=5$, $d=0$. A constant sequence is the flat, degenerate AP. It's still legal because "same difference" is satisfied ($d=0$ is a number).

$a=3$ (the first entry). $d=8-3=5$ (any consecutive difference). Use $a_n=a+(n-1)d$. Why $(n-1)$ and not $n$? Term 1 costs zero jumps — you're already standing on it. Reaching term 8 needs $8-1=7$ jumps. $a_8=3+(8-1)\cdot 5=3+35=38$

$a=4$, $d=3$. We know the value ($a_n=100$) and want the position $n$, so we solve $a_n=a+(n-1)d$ for $n$: $100=4+(n-1)\cdot 3\Rightarrow 96=3(n-1)\Rightarrow n-1=32\Rightarrow n=33.$ So $100$ is the $33$rd term. Sanity check: $n=33$ is a positive whole number, so $100$ genuinely appears in the list. Had we got a fraction, the answer would be "$100$ is not a term."

$a=2$, $d=3$, $n=30$. Use the standard sum form (best when you know $a$ and $d$ but not the last term yet): $S_{30}=\frac{30}{2}[2(2)+(30-1)\cdot 3]=15[4+87]=15\times 91=1365.$ Why this form? The Gauss Summation Trick pairs first-with-last, and $2a+(n-1)d$ is exactly $a_1+a_n$ written using only $a$ and $d$.

Here we already know $a=6$ and $a_{25}=54$, so the average form is fastest — no need to find $d$: $S_{25}=\frac{n}{2}(a+a_n)=\frac{25}{2}(6+54)=\frac{25}{2}\times 60=25\times 30=750.$ Read it as (number of terms) $\times$ (average term). The average of an evenly-spaced list is just $\frac{\text{first}+\text{last}}{2}=30$, which links to Arithmetic Mean.

Write each fact with $a_n=a+(n-1)d$: $a_4=a+3d=11,a_9=a+8d=26.$ Why subtract the equations? Subtracting kills $a$ and leaves only $d$: $(a+8d)-(a+3d)=26-11\Rightarrow 5d=15\Rightarrow d=3.$ Back-substitute: $a+3(3)=11\Rightarrow a=2$. Then $a_{15}=2+(15-1)\cdot 3=2+42=44.$

Use $a_n=S_n-S_{n-1}$. Why? $S_n$ counts terms $1$ through $n$; $S_{n-1}$ counts $1$ through $n-1$; the leftover is exactly the $n$th term. $a_n=(2n^2+3n)-(2(n-1{)}^2+3(n-1)).$ Expand $(n-1{)}^2=n^2-2n+1$: $a_n=2n^2+3n-(2n^2-4n+2+3n-3)=2n^2+3n-(2n^2-n-1)=4n+1.$ So $a_1=5$, $a_2=9$, $a_3=13$. Common difference $d=a_2-a_1=4$. First three terms: $5,9,13$.

The figure below plots these terms as dots against their position $n$. Notice the dots fall exactly on the straight lavender line $a_n=4n+1$: the green arrow marks one step of size $d=4$ between consecutive terms. The whole point of the picture is visual — because the term-values line up on a straight line, the jump between neighbours is the same everywhere, and that constant jump is what makes it an AP (linking to Linear Functions).

Compute the general term: $a_n=S_n-S_{n-1}=(p{n}^2+qn)-(p(n-1{)}^2+q(n-1)).$ $(n-1{)}^2=n^2-2n+1$, so $a_n=p{n}^2+qn-(p{n}^2-2pn+p+qn-q)=2pn-p+q.$ This is linear in $n$ (of the form $mn+c$). A linear $a_n$ is exactly the shape $a+(n-1)d$, which links to Linear Functions. The step between terms is $d=a_{n+1}-a_n=2p(n+1)-(2pn)=2p.$ Because $d$ came out as a constant (no $n$ left), the difference is the same every step ⇒ AP. (The "no constant term" condition matters: it guarantees $S_0=0$, i.e. the sum of zero terms is zero, as it should be.)

$a=9$, $d=3$, $S_n=189$. Since we know $a,d$ but not $n$, plug into the standard sum form and solve for $n$: $189=\frac{n}{2}[2(9)+(n-1)\cdot 3]=\frac{n}{2}[18+3n-3]=\frac{n}{2}(3n+15).$ Clear the fraction — why? to turn this into a plain quadratic we can factor: $378=n(3n+15)=3n^2+15n\Rightarrow 3n^2+15n-378=0.$ Divide the whole equation by $3$ (every coefficient is a multiple of 3) to make the numbers small: $n^2+5n-126=0\Rightarrow (n+14)(n-9)=0\Rightarrow n=-14\text{ or }n=9.$ Why reject $n=-14$? $n$ counts terms, so it must be a positive whole number. That leaves $n=9$. Always check the discriminant is a perfect square before trusting an answer: here $5^2+4(126)=25+504=529=23^2$, a perfect square, so a whole-number $n$ exists. If it hadn't been a perfect square, no number of terms would hit the target and the honest answer would be "unreachable." (This perfect-square check is verified in the =VERIFY= block at the bottom of this page.)

The five terms are $4,\_,\_,\_,20$: so $a_1=4$ and $a_5=20$. Use $a_5=a+4d$: $20=4+4d\Rightarrow 4d=16\Rightarrow d=4.$ Why 4 gaps, not 5? Five terms have exactly $5-1=4$ jumps between the ends. The inserted numbers are $4+4=8,8+4=12,12+4=16.$ The three numbers are $8,12,16$. (Inserting evenly-spaced terms is the many-term version of the Arithmetic Mean.)

Strategy: sum from 5 to 12 $=S_{12}-S_4$ (everything up to 12, minus everything up to 4). $a=3$, $d=4$. $S_{12}=\frac{12}{2}[2(3)+11\cdot 4]=6[6+44]=6\times 50=300.$ $S_4=\frac{4}{2}[2(3)+3\cdot 4]=2[6+12]=2\times 18=36.$ Slice sum $=300-36=264$. Cross-check with average form: the slice has $12-5+1=8$ terms, first $=a_5=3+16=19$, last $=a_{12}=3+44=47$; sum $=\frac{8}{2}(19+47)=4\times 66=264$ ✓.

Turn each condition into an equation using $S_n=\frac{n}{2}[2a+(n-1)d]$. $S_7=\frac{7}{2}(2a+6d)=49\Rightarrow 2a+6d=14\Rightarrow a+3d=7.$ $S_{17}=\frac{17}{2}(2a+16d)=289\Rightarrow 2a+16d=34\Rightarrow a+8d=17.$ Subtract: $5d=10\Rightarrow d=2$, then $a=7-3(2)=1$. $S_n=\frac{n}{2}[2(1)+(n-1)\cdot 2]=\frac{n}{2}(2n)={\mathbf{n}}^2.$ Neat: the sum of the first $n$ odd numbers ($1,3,5,\dots$) is $n^2$. Verify: $S_7=49$ ✓, $S_{17}=289$ ✓.

Key idea: the $m$th term of an AP equals the average form run backwards — specifically $a_m=\frac{S_{2m-1}}{2m-1}$, because $S_{2m-1}=(2m-1)\times (\text{middle term})=(2m-1)a_m$. Why? In an AP, the average of the first $2m-1$ terms is the middle term $a_m$, and (number of terms)×(average) = sum. For the 12th term set $m=12\Rightarrow 2m-1=23$. So $\frac{a_{12}^{(1)}}{a_{12}^{(2)}}=\frac{S_{23}^{(1)}}{S_{23}^{(2)}}=\frac{3(23)+8}{7(23)+15}=\frac{69+8}{161+15}=\frac{77}{176}=\frac{7}{16}.$ The 12th terms are in ratio $7:16$.

(a) Terms: $a_n=20+(n-1)(-3)=23-3n$. Set $a_n<0$: $23-3n<0\Rightarrow n>7.67$, so the 8th term is the first negative one ($a_7=2>0$, $a_8=-1<0$). (b) The partial sum grows while terms are positive and shrinks once they turn negative. So the peak is reached after adding the last positive term, term 7. $S_7=\frac{7}{2}[2(20)+6(-3)]=\frac{7}{2}(40-18)=\frac{7}{2}\times 22=77.$ Check the neighbours: $S_8=S_7+a_8=77+(-1)=76<77$ ✓, and $S_6=75<77$ ✓. The maximum is $77$ at $n=7$. The figure shows the sum climbing then falling.