Arithmetic Progressions — RBSE Class 10 (Mathematics)

Stack chairs, save a fixed amount each month, count the seats in widening rows of an auditorium — again and again life throws up lists where each step adds the same amount. Such a list is an Arithmetic Progression, and two neat formulas let you jump to the 100th term or add up the first fifty without writing them all out.

1. What makes a sequence an AP

An AP is a list of numbers in which each term is obtained by adding a fixed number, the common difference d, to the previous term. $a,a+d,a+2d,a+3d,\dots$

• $a$ = first term.
• $d=a_2-a_1=a_3-a_2$ = common difference (can be positive, negative or zero).

Test for an AP: the difference between consecutive terms must be constant. For 3, 7, 11, 15… the difference is always 4 → it is an AP with $a=3,d=4$.

2. The nth term

$\boxed{a_n=a+(n-1)d}$

This is the workhorse formula. It lets you find any term directly, or — read backwards — find $n$, $a$ or $d$ when the rest are known.

Example — find the 10th term of 2, 7, 12, … $a=2,d=5\Rightarrow a_{10}=2+9(5)=47$.

The nth term from the end of an AP with last term $l$ is $l-(n-1)d$.

3. Sum of the first n terms

$\boxed{S_n=\frac{n}{2}[2a+(n-1)d]}$

If the last term $l=a_n$ is known, a shorter form is handy: $S_n=\frac{n}{2}(a+l)$

A useful link: $a_n=S_n-S_{n-1}$ — the nth term is the jump in the running total.

Example — sum of the first 20 terms of 1, 3, 5, … (odd numbers). $a=1,d=2\Rightarrow S_{20}=\frac{20}{2}[2+19(2)]=10\times 40=400$. (Indeed the sum of the first n odd numbers is $n^2$.)

4. Typical exam manoeuvres

• "Which term is …?" Set $a_n=$ value, solve for n. If n is not a positive integer, that value is not a term.
• Given two terms, form two equations in $a$ and $d$ and solve (an AP problem hiding a linear pair).
• Three terms in AP: taking them as $a-d,a,a+d$ makes the sum instantly $3a$ — a great time-saver.
• Sum given, find n: $S_n$ is quadratic in n, so you may get two values — discard any that is negative or non-integer.

5. Worked setup — the auditorium

An auditorium has 20 seats in the first row and 2 more in each successive row; there are 30 rows. Total seats? $a=20,d=2,n=30$: $S_{30}=\frac{30}{2}[2(20)+29(2)]=15[40+58]=15\times 98=1470$ seats.

6. Closing thought

Everything reduces to two formulas — $a_n=a+(n-1)d$ and $S_n=\frac{n}{2}[2a+(n-1)d]$ — plus the habit of first pinning down $a$ and $d$. In the RBSE board this chapter is generous: an easy "find the term/sum" question and a longer word problem appear almost every year, and the arithmetic is straightforward once the two formulas are automatic.