Arithmetic sequences and series — IB Maths AA practice questions

Question 1

In an arithmetic progression, $a_3 = 11$ and $a_7 = 27$. Determine $a_5$.

Accepted Answer

$a_5 = \dfrac{a_3 + a_7}{2} = \dfrac{11 + 27}{2} = 19$.

Answer

$15$

Answer

$17$

Answer

$19$

Answer

$22$

Question 2

In an arithmetic progression with $a_1 = 3$ and $r = 4$, the term $a_n = 47$. Determine $n$.

Accepted Answer

$47 = 3 + (n-1) \cdot 4 \Rightarrow (n-1) \cdot 4 = 44 \Rightarrow n - 1 = 11 \Rightarrow n = 12$.

Answer

$10$

Answer

$11$

Answer

$12$

Answer

$13$

Question 3

A student deposits $\$50$ in January and increases the deposit by $\$10$ every following month. How much will be deposited in total over $12$ months?

Accepted Answer

$S_{12} = \dfrac{12\,(2 \cdot 50 + 11 \cdot 10)}{2} = 6 \cdot 210 = 1260$.

Answer

$\$960$

Answer

$\$1{,}080$

Answer

$\$1{,}260$

Answer

$\$1{,}320$

Question 4

Compute the sum of the first $20$ terms of the arithmetic progression $(a_n)_{n \ge 1}$ with $a_1 = 2$ and $r = 3$.

Accepted Answer

$S_{20} = \dfrac{20\,(2 \cdot 2 + 19 \cdot 3)}{2} = \dfrac{20 \cdot 61}{2} = 10 \cdot 61 = 610$.

Answer

$560$

Answer

$590$

Answer

$610$

Answer

$620$

Question 5

For the arithmetic progression with $a_1 = 1$ and $r = 2$, compute $a_5 + a_6 + a_7 + a_8 + a_9$.

Accepted Answer

$a_n = 2n - 1$, so $a_5 + \cdots + a_9 = 9 + 11 + 13 + 15 + 17 = 65$. Equivalently, this is the sum of $5$ consecutive odd numbers starting at $9$.

Answer

$35$

Answer

$45$

Answer

$65$

Answer

$75$

Question 6

The numbers $x - 2$, $5$, $x + 4$ are in arithmetic progression (in this order). Determine $x$.

Accepted Answer

$5 = \dfrac{(x - 2) + (x + 4)}{2} \Rightarrow 10 = 2x + 2 \Rightarrow x = 4$.

Answer

$1$

Answer

$2$

Answer

$4$

Answer

$7$

Question 7

In the arithmetic progression $(a_n)_{n \ge 1}$, $a_2 = 7$ and $a_5 = 16$. Determine $a_1$.

Accepted Answer

$r = \dfrac{16 - 7}{3} = 3$, so $a_1 = a_2 - r = 7 - 3 = 4$.

Answer

$3$

Answer

$4$

Answer

$5$

Answer

$7$

Question 8

The numbers $5, 8, 11, 14, \ldots$ form an arithmetic progression. The common difference $r$ equals:

Accepted Answer

$r = a_2 - a_1 = 8 - 5 = 3$.

Answer

$2$

Answer

$3$

Answer

$4$

Answer

$5$

Question 9

For an arithmetic progression with $a_1 = 5$ and $r = 3$, find the smallest $n$ such that $S_n \ge 500$.

Accepted Answer

$S_n = \dfrac{n(3n + 7)}{2}$. Compute: $S_{17} = \dfrac{17 \cdot 58}{2} = 493 < 500$ and $S_{18} = \dfrac{18 \cdot 61}{2} = 549 \ge 500$. Hence the smallest $n$ is $18$.

Answer

$16$

Answer

$17$

Answer

$18$

Answer

$19$

Question 10

Three numbers in arithmetic progression have sum $15$ and the sum of their squares is $83$. The largest of them is:

Accepted Answer

Set the terms as $5 - r,\, 5,\, 5 + r$. Then $(5-r)^2 + 25 + (5+r)^2 = 50 + 2r^2 + 25 = 83$, so $2r^2 = 8 \Rightarrow r = 2$. The largest term is $5 + 2 = 7$.

Answer

$5$

Answer

$6$

Answer

$7$

Answer

$8$

Arithmetic sequences and series practice questions — IB Maths AA

Problems & worked solutions

Problem 1 — Arithmetic sequences and series

Problem 2 — Arithmetic sequences and series

Problem 3 — Arithmetic sequences and series

Problem 4 — Arithmetic sequences and series

Problem 5 — Arithmetic sequences and series

Problem 6 — Arithmetic sequences and series

Problem 7 — Arithmetic sequences and series

Problem 8 — Arithmetic sequences and series

Problem 9 — Arithmetic sequences and series

Problem 10 — Arithmetic sequences and series