4.5 Arithmetic sequences and series — Practise with help

1. $u_5 = 3(5) + 2 = 17$.
2. $\sum_{r=1}^{4}r = \frac{4 \times 5}{2} = 10$.
3. Subtract $3$: $2x = 12$, so $x = 6$.

If question 2 was hard, revisit the standard results in Section 4.4 before you continue. The series formula in this section comes from the same idea.

Find the first term $a$ and the common difference $d$ first, then use $u_n = a + (n - 1)d$.

Here $a = 4$ and $d = 3$. Substitute $n = 12$ into the formula, so the number of steps is $11$.

The 12th term is $37$. The number of steps is $11$, which is one less than the position.

The sequence falls, so the common difference is negative. Work out $17 - 20$.

Here $a = 20$ and $d = -3$. Substitute $n = 10$ into $u_n = a + (n - 1)d$.

The 10th term is $-7$. The sequence crosses zero between the 7th and 8th terms, so a negative term further along is expected.

Use $S_n = \dfrac{n}{2}\bigl(2a + (n - 1)d\bigr)$ with $a = 6$ and $d = 4$.

Substitute $n = 15$: the bracket is $2(6) + 14(4) = 12 + 56$. Then multiply by $\frac{15}{2}$.

The sum of the first 15 terms is $510$. The bracket already includes the $14$ steps, so no further adjustment is needed.

Find how many terms there are before you add them. Set the $n$th term equal to $81$.

Solve $5 + (n - 1)(4) = 81$. Then use $S_n = \dfrac{n}{2}(a + l)$ with the last term $l = 81$.

Set the term equal to the last term:

There are 20 terms. Then

The sum is $860$.

The income is arithmetic with $a = 150$ and $d = 50$. The question asks for a total, so use the series formula rather than the term formula.

Substitute $n = 8$: the bracket is $2(150) + 7(50) = 300 + 350$.

The quiz night raises £2600 in the first 8 months. The eighth month alone raises $150 + 7(50) = 500$ pounds, which is less than a quarter of the total.

Each term is built from $a$ and $d$. Write two equations, one for the 5th term and one for the 10th.

The two equations are $a + 4d = 17$ and $a + 9d = 37$. Subtract the first from the second to remove $a$.

Subtracting gives

Substitute $d = 4$ into the first equation: $a + 16 = 17$, so $a = 1$. The sequence is $1, 5, 9, 13, \dots$, and the 5th term is $1 + 16 = 17$, which checks.

List the first few: $7, 14, 21, \dots$ The last multiple below $100$ is $98$. The series is arithmetic.

Find $n$ from $7n = 98$, then use $S_n = \frac{n}{2}(7 + 98)$.

The multiples are $7, 14, 21, \dots, 98$. There are $\frac{98}{7} = 14$ of them. Then

The sum of all the multiples of $7$ below $100$ is $735$. Every term is $7$ times a whole number, so the common difference is $7$.