Worked examples · Sheet 2

Problems with arithmetic progressions

Theory used: Arithmetic progression Sum of the terms of an arithmetic progression

Exercise 1

In an arithmetic progression we have $a_3 = 11$ and $a_8 = 26$ . Find the common difference, the first term, the twentieth term and the sum of the first twenty terms.

a_8 - a_3 = 5d \iff 26 - 11 = 5d \iff d = 3

a_1 = a_3 - 2d = 11 - 6 = 5

a_{20} = a_1 + 19d = 5 + 57 = 62

S_{20} = \frac{20\,(a_1 + a_{20})}{2} = 10 \cdot 67 = 670

From

a_{3} = 11

a_{8} = 26

there are

5

steps and the change is

15

, so the common difference is

d = \frac{15}{5} = 3

In an arithmetic progression, to go from $a_3$ to $a_8$ one adds the common difference five times, so the difference $a_8 - a_3$ equals $5d$ , and $d = 3$ . The first term is obtained by going back two positions from $a_3$ , that is, by subtracting the common difference twice. The twentieth term is found with the formula $a_n = a_1 + (n - 1)\,d$ . For the sum of the terms, one multiplies the number of terms by the mean of the first and the last, and obtains $670$ .

Exercise 2

Consider the sum $7 + 11 + 15 + 19 + \dots$ . Find how many terms are needed for the sum to exceed $1000$ .

a_1 = 7, \qquad d = 4, \qquad a_n = 7 + 4\,(n - 1) = 4n + 3

S_n = \frac{n\,(7 + 4n + 3)}{2} = \frac{n\,(4n + 10)}{2} = n\,(2n + 5)

2n^2 + 5n > 1000 \iff n > \frac{-5 + \sqrt{25 + 8000}}{4} \approx 21.15

S_{21} = 21 \cdot 47 = 987, \qquad S_{22} = 22 \cdot 49 = 1078

The terms increase by $4$ each time, so they form an arithmetic progression with first term $7$ and common difference $4$ . With the formula for the sum of the terms, the sum of the first $n$ terms becomes the polynomial $n\,(2n + 5)$ . The condition $S_n > 1000$ is a quadratic inequality in $n$ ; the parabola $2n^2 + 5n - 1000$ opens upward, so it is positive beyond the positive root, about $21.15$ , and the negative root is irrelevant because $n$ is a natural number. The first natural number beyond $21.15$ is $22$ , therefore 22 terms are needed, as confirmed by $S_{21} = 987$ and $S_{22} = 1078$ .

Exercise 3

Find three numbers in increasing arithmetic progression whose sum is $21$ and whose product is $231$ .

x - d, \qquad x, \qquad x + d

(x - d) + x + (x + d) = 3x = 21 \iff x = 7

(7 - d) \cdot 7 \cdot (7 + d) = 231 \iff 49 - d^2 = 33 \iff d^2 = 16

d > 0 \implies d = 4: \qquad 3,\ 7,\ 11

Three numbers in arithmetic progression are written as the middle term $x$ preceded and followed by the common difference $d$ , that is, $x - d$ , $x$ , $x + d$ . With this choice the common difference cancels in the sum, and the first condition immediately gives $x = 7$ . The product contains the difference of squares $(7 - d)(7 + d) = 49 - d^2$ , so dividing by $7$ gives $d^2 = 16$ . Since the progression is increasing the common difference is positive, therefore $d = 4$ and the three numbers sought are $3$ , $7$ and $11$ . As a check, $3 + 7 + 11 = 21$ and $3 \cdot 7 \cdot 11 = 231$ .