Arithmetic Sequences and Series: Practice

The common difference is what you add to move from any term to the next, so subtract a term from the one after it.

$11 - 7 = 4, \quad 15 - 11 = 4, \quad 19 - 15 = 4$

Every difference agrees, so $d = 4$. The $7$ is the first term, not the difference.

Getting from the 1st term to the 4th takes $4 - 1 = 3$ steps, and each step adds $5$.

$a_4 = a_1 + (4-1)d = 3 + 3 \cdot 5 = 18$

Multiplying by $4$ instead of $3$ gives $23$, which is really the 5th term. Count the steps, not the terms.

From the 1st term to the 10th is $10 - 1 = 9$ steps.

$a_{10} = 2 + 9 \cdot 6 = 2 + 54 = 56$

Taking $10$ steps instead would give $2 + 60 = 62$, the 11th term. The first term costs no steps at all.

Here $a_1 = 4$ and $d = 9 - 4 = 5$, so substitute into $a_n = a_1 + (n-1)d$ and simplify.

$a_n = 4 + 5(n-1) = 4 + 5n - 5 = 5n - 1$

Test it against the sequence: $a_1 = 5 - 1 = 4$ and $a_2 = 10 - 1 = 9$, both correct.

The common difference is $4$, and the climb from the first term to the last is $43 - 3 = 40$, so there are $40 \div 4 = 10$ steps.

$n = \frac{43 - 3}{4} + 1 = 10 + 1 = 11$

Steps are not terms. You are already standing on a term before you take a single step, so the number of terms is one more than the number of steps.

First find the 6th term. From the 1st to the 6th is $5$ steps of $d = 4$.

$a_6 = 3 + 5 \cdot 4 = 23$

Now add the six terms with the sum formula.

$S_6 = \frac{6}{2}(3 + 23) = 3 \cdot 26 = 78$

Forgetting to halve gives $156$, which is $2S_6$, the total of the forwards and backwards copies together.

The whole numbers from $1$ to $100$ are arithmetic with $a_1 = 1$, $a_{100} = 100$, and $n = 100$.

$S_{100} = \frac{100}{2}(1 + 100) = 50 \cdot 101 = 5050$

Written forwards and backwards, the sum has $100$ columns each totalling $101$, so $2S = 10{,}100$ and the sum itself is $5{,}050$.

Reaching the 7th term from the 1st takes $6$ steps.

$a_7 = -4 + 6 \cdot 3 = -4 + 18 = 14$

Using $7$ steps gives $17$, which is the 8th term. Subtracting instead of adding gives $-22$.

The common difference is $d = 9 - 5 = 4$, and reaching the 12th term takes $11$ steps.

$a_{12} = 5 + 11 \cdot 4 = 5 + 44 = 49$

The usual slip is $5 + 12 \cdot 4 = 53$, which overshoots by one full step.

The rule says every term is $2$ less than the one before it, so the sequence is arithmetic with $d = -2$. Reaching the 5th term from the 1st takes $4$ steps.

$a_5 = 6 + 4(-2) = 6 - 8 = -2$

Listing them confirms it: $6, 4, 2, 0, -2$.

The last two terms reveal the common difference: $26 - 20 = 6$.

$8 + 6 = 14, \quad 14 + 6 = 20$

So the missing term is $14$, and now every consecutive difference is $6$. As a check, the middle of three arithmetic terms is the average of its neighbours: $\frac{8 + 20}{2} = 14$.

First count the terms, using $a_1 = 4$, $d = 3$, and last term $28$.

$n = \frac{28 - 4}{3} + 1 = 8 + 1 = 9$

Now add the nine terms.

$S_9 = \frac{9}{2}(4 + 28) = \frac{9}{2} \cdot 32 = 9 \cdot 16 = 144$

Forgetting the $+1$ and using $n = 8$ would give $128$, and forgetting the half would give $288$.