Arithmetic Sequences and Series: Practice

The rule is linear in $k$, so the terms are arithmetic with $d = -3$. Find the two ends.

$a_1 = 100 - 3 = 97, \qquad a_{40} = 100 - 120 = -20$

There are $40$ terms, so apply the sum.

$S_{40} = \frac{40}{2}(97 - 20) = 20 \cdot 77 = 1540$

The terms turn negative partway through, which changes nothing: the pairing argument never assumed the terms were positive.

The first sequence steps by $7$, so its terms are $2, 9, 16, 23, 30, \ldots$. The second steps by $5$, so its terms are $3, 8, 13, 18, 23, 28, \ldots$.

Walk along the first list and test each entry against the second.

$2, \; 9, \; 16, \; \mathbf{23} \qquad \text{and} \qquad 3, \; 8, \; 13, \; 18, \; \mathbf{23}$

The first value they share is $23$. Note that $16$ and $30$ belong only to the first list, and $18$ only to the second.

Consecutive terms means the two differences are the same, so $b - a = c - b$.

Add $a + b$ to both sides of that equation.

$b - a = c - b \;\Rightarrow\; 2b = a + c$

The middle term is therefore the average of its neighbours, which is exactly what an arithmetic mean is. The other three statements all fail on a plain example such as $5, 8, 11$, where $a + c = 16$ but $b = 8$, $a + b = 13$ but $2c = 22$, and $c - a = 6$ but $b = 8$.

The list $51, 52, \ldots, 100$ is arithmetic with $d = 1$, so count the terms first.

$n = \frac{100 - 51}{1} + 1 = 50$

Now add them.

$S = \frac{50}{2}(51 + 100) = 25 \cdot 151 = 3775$

A second route confirms it: the sum to $100$ minus the sum to $50$ is $5{,}050 - 1{,}275 = 3{,}775$.

The smallest three-digit multiple of $7$ is $105 = 7 \cdot 15$ and the largest is $994 = 7 \cdot 142$, and they form an arithmetic sequence with $d = 7$.

$n = \frac{994 - 105}{7} + 1 = \frac{889}{7} + 1 = 127 + 1 = 128$

The $127$ counts the steps between the ends, and the $+1$ counts the term you started on. Checking the multipliers agrees: they run from $15$ to $142$, and $142 - 15 + 1 = 128$.

The running total grows as long as the terms being added are positive, and shrinks once they go negative. So find the last positive term.

$a_n = 20 + (n-1)(-3) = 23 - 3n$

This stays positive while $n < \frac{23}{3} \approx 7.67$, so $a_7 = 2$ is the last positive term and $a_8 = -1$ is the first negative one.

Adding $a_8$ would pull the total down, so the maximum is at $n = 7$, where $S_7 = \frac{7}{2}(20 + 2) = 77$ (against $S_8 = 76$).

With three numbers inserted, the finished sequence has $5$ terms, so there are $4$ steps from the first to the last.

$d = \frac{24 - 4}{5 - 1} = \frac{20}{4} = 5$

Now step up from $4$ by fives.

$4, \; 9, \; 14, \; 19, \; 24$

Dividing by $5$ (the number of terms) instead of $4$ (the number of steps) would give $d = 4$ and the wrong list $8, 12, 16$.

The terms are arithmetic with $a_1 = 5$ and last term $a_n = 3n + 2$, so the sum is

$S_n = \frac{n}{2}\big(5 + (3n + 2)\big) = \frac{n(3n + 7)}{2}.$

Set it equal to $185$ and clear the fraction.

$3n^2 + 7n - 370 = 0 \;\Rightarrow\; (n - 10)(3n + 37) = 0$

The only positive whole number is $n = 10$, and it checks: $3(1 + 2 + \cdots + 10) + 2(10) = 165 + 20 = 185$.

The two given terms sit $p - q$ steps apart, so start there.

$d = \frac{a_p - a_q}{p - q} = \frac{\frac{1}{q} - \frac{1}{p}}{p - q} = \frac{\frac{p - q}{pq}}{p - q} = \frac{1}{pq}$

Now back out the first term, which is $p - 1$ steps below the $p$th.

$a_1 = \frac{1}{q} - \frac{p-1}{pq} = \frac{p - (p - 1)}{pq} = \frac{1}{pq}$

So $a_1 = d = \frac{1}{pq}$, which makes $a_n = \frac{n}{pq}$ and in particular $a_{pq} = 1$. The sum follows.

$S_{pq} = \frac{pq}{2}\left(\frac{1}{pq} + 1\right) = \frac{1 + pq}{2}$

The condition $p, q > 1$ matters more than it looks: $\frac{p + q}{2}$ agrees with the answer exactly when $(p-1)(q-1) = 0$, so allowing $p = 1$ would have made two of these choices equal.

In an arithmetic sequence the total $a_i + a_j$ depends only on $i + j$, because

$a_i + a_j = \big(a_1 + (i-1)d\big) + \big(a_1 + (j-1)d\big) = 2a_1 + (i + j - 2)d.$

Here $2 + 14 = 16$ and $8 + 8 = 16$, so the pair $(a_2, a_{14})$ has the same total as $(a_8, a_8)$.

$2a_8 = 30 \;\Rightarrow\; a_8 = 15$

This is the same identity that makes every column of the forwards-and-backwards sum come out equal.

Name the terms $m - d$, $m$, and $m + d$, centring them on the middle one. The sum then collapses.

$(m - d) + m + (m + d) = 3m = 27 \;\Rightarrow\; m = 9$

The outer two multiply to $m^2 - d^2$, so the product gives a second equation.

$9(81 - d^2) = 693 \;\Rightarrow\; 81 - d^2 = 77 \;\Rightarrow\; d^2 = 4$

So $d = 2$ or $d = -2$, and either way the three terms are $7$, $9$, and $11$. The largest is $11$.

Write both sums in the form built from $a_1$ and $d$.

$S_5 = \frac{5}{2}(2a_1 + 4d) = 5(a_1 + 2d) = 40, \qquad S_{10} = 5(2a_1 + 9d) = 155$

The first says $a_1 + 2d = 8$ and the second says $2a_1 + 9d = 31$. Substitute $a_1 = 8 - 2d$ into the second.

$2(8 - 2d) + 9d = 31 \;\Rightarrow\; 16 + 5d = 31 \;\Rightarrow\; d = 3$

Then $a_1 = 8 - 2(3) = 2$. Check both: $S_5 = 5(2 + 6) = 40$ and $S_{10} = 5(4 + 27) = 155$.