Arithmetic Progressions: Important Questions (Part 2)

This lesson revises the main formulae for an arithmetic progression (AP), then works through four exam-style questions: two on counting terms in a range, one on finding the first negative term, and one on finding the middle term from a sum.

Key formulae for an arithmetic progression

The general form of an AP with first term $a$ and common difference $d$ is

$a, \ a + d, \ a + 2d, \ \dots$

The common difference is the gap between consecutive terms:

$d = a_2 - a_1 = a_3 - a_2 = a_n - a_{n-1}.$

The nth term is

$a_n = a + (n - 1)d.$

The sum of the first $n$ terms is

$S_n = \frac{n}{2}\bigl(2a + (n - 1)d\bigr) = \frac{n}{2}(a + a_n).$

The number of terms, when the first and last terms are known, is

$n = \frac{a_n - a_1}{d} + 1.$

A single term can also be recovered from sums as $a_n = S_n - S_{n-1}$. If $a$, $b$, $c$ are in AP, the middle term is the average of the outer two:

$b = \frac{a + c}{2}.$

It is also useful to take consecutive terms symmetrically: three as $a - d,\ a,\ a + d$, and four as $a - 3d,\ a - d,\ a + d,\ a + 3d$.

Counting integers divisible by 5 between 220 and 700

We want the integers between $220$ and $700$ that are exactly divisible by $5$. The first such number after $220$ is

$a_1 = 225,$

and the last such number before $700$ is

$a_n = 695, \qquad d = 5.$

Using the count formula:

$n = \frac{a_n - a_1}{d} + 1 = \frac{695 - 225}{5} + 1 = \frac{470}{5} + 1 = 94 + 1 = 95.$

Answer: there are $95$ such integers.

Counting integers divisible by 2 from 100 to 900

We want the integers from $100$ to $900$ that are exactly divisible by $2$. Since $100$ is itself even, it is included:

$a_1 = 100, \qquad a_n = 900, \qquad d = 2.$

Then

$n = \frac{a_n - a_1}{d} + 1 = \frac{900 - 100}{2} + 1 = \frac{800}{2} + 1 = 400 + 1 = 401.$

Answer: there are $401$ such integers.

First negative term of an AP

For the AP $12, \ 10, \ 8, \ \dots$ we have

$a = 12, \qquad d = 10 - 12 = -2.$

The first negative term is the smallest $n$ with $a_n < 0$:

$a + (n - 1)d < 0$ $12 + (n - 1)(-2) < 0$ $12 - 2n + 2 < 0$ $14 - 2n < 0$ $-2n < -14.$

Dividing by $-2$ reverses the inequality:

$n > 7.$

The smallest integer greater than $7$ is $8$, so the first negative term is the 8th term. Check: $a_8 = 12 + 7(-2) = -2 < 0$.

Middle term from three consecutive terms

Three consecutive terms of an AP have sum $27$; find the middle term. Take the terms symmetrically as $a - d,\ a,\ a + d$. Then

$(a - d) + a + (a + d) = 27.$

The $-d$ and $+d$ cancel, leaving

$3a = 27 \implies a = 9.$

The middle term is $a$, so the middle term is $9$.

Key takeaways

• To count terms in a range divisible by a number, find the first and last valid values and use $n = \dfrac{a_n - a_1}{d} + 1$.
• The first negative term comes from solving $a + (n-1)d < 0$; dividing by a negative number reverses the inequality.
• Writing three consecutive terms as $a - d,\ a,\ a + d$ makes their sum collapse to $3a$, so the middle term is the sum divided by $3$.