Why Adding a Pattern Forwards and Backwards Gives Its Sum in One Line

Write the sum of an AP forwards, then backwards underneath:

$$S=a+(a+d)+\cdots +(a+(n-1)d)$$
$$S=(a+(n-1)d)+(a+(n-2)d)+\cdots +a$$

**Each column adds to the same total, $2a+(n-1)d$**, and there are $n$ columns. So $2S=n[2a+(n-1)d]$.

This part covers both sum formulas, finding the number of terms, getting a term from sums, and real-life problems.

**Substitute the first term $a$, common difference $d$ and number of terms $n$ into $S_n=\frac{n}{2}[2a+(n-1)d]$.

Worked example 1.** Sum of the first $20$ terms of $3,7,11,\dots$

$$S_{20}=\frac{20}{2}[6+19\times 4]=10\times 82=820$$

Worked example 2. Sum of the first $15$ terms of $10,7,4,\dots$

$$S_{15}=\frac{15}{2}[20+14\times (-3)]=7.5\times (-22)=-165$$

Worked example 3. Sum of the first $50$ natural numbers:

$$S_{50}=\frac{50}{2}[2+49]=25\times 51=1275$$

An everyday example. **A school auditorium has $20$ rows; the front row has $20$ seats and each row behind has $2$ more.** Total seats:

$$S_{20}=10[40+19\times 2]=10\times 78=780$$

The substance. A sum can be negative when enough terms are negative, as in the second example.

**Find the number of terms $n$ if it is not given, then use $S_n=\frac{n}{2}(a+l)$, where $l$ is the last term.**

This is the same formula, since $l=a+(n-1)d$.

Worked example 1. Find $7+13+19+\cdots +205$. This AP has $34$ terms.

$$S=\frac{34}{2}(7+205)=17\times 212=3604$$

Worked example 2. Sum of all two-digit multiples of $7$: $14,21,\dots ,98$, which is $13$ terms.

$$S=\frac{13}{2}(14+98)=\frac{13\times 112}{2}=728$$

Worked example 3. Sum of the odd numbers from $1$ to $99$: there are $50$ of them.

$$S=\frac{50}{2}(1+99)=2500$$

An everyday example. **A chapter in a textbook runs from page $45$ to page $72$.** That is $28$ pages, and the page numbers add to $\frac{28}{2}(45+72)=14\times 117=1638$.

The substance. **The sum of the first $n$ odd numbers is always $n^2$** — here $50^2=2500$.

**Set $S_n$ equal to the target and solve the quadratic in $n$, keeping only positive whole numbers; to find the $n$th term from a formula for $S_n$, use $a_n=S_n-S_{n-1}$.

Worked example 1.** How many terms of $24,21,18,\dots$ give a sum of $78$?

$$\frac{n}{2}[48+(n-1)(-3)]=78n(51-3n)=156n^2-17n+52=0(n-4)(n-13)=0$$

**Both $n=4$ and $n=13$ work**, because terms $5$ to $13$ are $12,9,6,3,0,-3,-6,-9,-12$, which add to zero.

Worked example 2. If $S_n=3n^2+5n$, find $a_n$.

$$a_n=S_n-S_{n-1}=3n^2+5n-[3(n-1{)}^2+5(n-1)]=6n+2$$

So $a_1=8$, $d=6$; check $S_2=22=8+14$.

An everyday example. Meera saves ₹50 in the first week, ₹60 in the second, and so on. When will she reach ₹1610?

$$\frac{n}{2}[100+10(n-1)]=1610n^2+9n-322=0(n+23)(n-14)=0n=14\text{ weeks}$$

The substance. Two positive answers can both be correct, so check each root instead of rejecting one automatically.

**Identify $a$, $d$ and what is being asked — a term, a sum or the number of terms — then apply the matching formula and interpret the answer in the situation.

Worked example 1 — instalments.** A loan of ₹3250 is repaid by paying ₹20 in the first month and ₹15 more each month. How many months does it take?

$$\frac{n}{2}[40+15(n-1)]=32503n^2+5n-1300=0$$
$$n=\frac{-5+\sqrt{25+15600}}{6}=\frac{-5+125}{6}=20\text{ months}$$

Worked example 2 — stacked logs. A pile of logs has $20$ in the bottom row, $19$ in the next, and so on up to $5$ in the top row.

$$n=20-5+1=16\text{ rows}S=\frac{16}{2}(20+5)=200\text{ logs}$$

Worked example 3 — savings. A child saves ₹5 on the first day and ₹3 more each day for $30$ days.

$$S_{30}=15[10+29\times 3]=15\times 97=₹1455$$

An everyday example. A timber depot stacking logs in rows uses exactly this reasoning to count its stock without counting each log.

The substance. **Reject negative or fractional values of $n$** — a number of months or rows must be a positive whole number.