Arithmetic Sequences and Series: Additional Practice (Set 2)

Write both sums with first term $a$ and common difference $d$.

$S_{10} = 5(2a + 9d), \qquad S_{20} = 10(2a + 19d)$

Setting them equal gives $2a + 9d = 4a + 38d$, which rearranges to $2a + 29d = 0$.

That combination is exactly what the next sum is built from.

$S_{30} = \frac{30}{2}(2a + 29d) = 15 \cdot 0 = 0$

So the first $30$ terms must total zero, whatever $a$ and $d$ happen to be.

From term $7$ to term $15$ is $8$ steps, over which the value falls by $-21 - 3 = -24$.

$d = \frac{-21 - 3}{15 - 7} = \frac{-24}{8} = -3$

Now step back the $6$ steps from the 7th term to the 1st. Undoing a step of $-3$ adds $3$.

$a_1 = 3 - 6(-3) = 3 + 18 = 21$

Check both given terms: $a_7 = 21 + 6(-3) = 3$ and $a_{15} = 21 + 14(-3) = -21$.

An arithmetic sum is the number of terms times the average of the first and last, and when the number of terms is odd, the middle term is that average.

$S_9 = \frac{9}{2}(a_1 + a_9) = 9 \cdot \frac{a_1 + a_9}{2} = 9a_5$

So divide.

$a_5 = \frac{189}{9} = 21$

The common difference never enters: whatever $d$ is, the middle term is pinned by the total.

The ten steps from term $10$ to term $20$ drop the value by $20$.

$d = \frac{20 - 40}{20 - 10} = -2$

Write the general term measuring from the 10th, which sits $n - 10$ steps away.

$a_n = 40 + (n - 10)(-2) = 60 - 2n$

Set it to zero: $60 - 2n = 0$ gives $n = 30$. Check it: $a_{30} = 40 + 20(-2) = 0$.

The two given terms are $n - m$ steps apart, so the common difference falls out at once.

$d = \frac{a_n - a_m}{n - m} = \frac{m - n}{n - m} = -1$

Now back out the first term, remembering that $m - 1$ steps separate it from $a_m$.

$a_1 = a_m - (m-1)d = n + (m - 1)$

Finally take $m + n - 1$ steps forward from there, each one subtracting $1$.

$a_{m+n} = (m + n - 1) - (m + n - 1) = 0$

The sequence steps down by $4$, so write the general term.

$a_n = 100 + (n-1)(-4) = 104 - 4n$

Positive means $104 - 4n > 0$, that is $n < 26$. The largest whole number that works is $n = 25$, and indeed $a_{25} = 4$.

$a_{26} = 104 - 104 = 0$

The 26th term is zero, and zero is not positive, so exactly $25$ terms are positive. That boundary is the whole trap in this question.

First get the common difference of the full sequence, which spans $19$ steps.

$d = \frac{60 - 3}{19} = 3$

The terms $a_1, a_3, \ldots, a_{19}$ skip every other one, so they are themselves an arithmetic sequence with common difference $2d = 6$, first term $3$, and last term $a_{19} = 3 + 18(3) = 57$. There are $10$ of them.

$\frac{10}{2}(3 + 57) = 5 \cdot 60 = 300$

As a check, all twenty terms sum to $\frac{20}{2}(3 + 60) = 630$, so the even-numbered terms account for the remaining $330$.

The three-digit multiples of $9$ run from $108 = 9 \cdot 12$ up to $999 = 9 \cdot 111$, stepping by $9$. Count them.

$n = \frac{999 - 108}{9} + 1 = 99 + 1 = 100$

Now add the hundred terms.

$S_{100} = \frac{100}{2}(108 + 999) = 50 \cdot 1107 = 55{,}350$

Forgetting the half gives $110{,}700$, exactly twice the answer.

The numbers from $1$ to $n$ are arithmetic with both ends known, so

$S_n = \frac{n}{2}(1 + n).$

Set that equal to $210$ and clear the fraction.

$n^2 + n - 420 = 0 \;\Rightarrow\; (n - 20)(n + 21) = 0$

A count cannot be negative, so $n = 20$, and it checks: $\frac{20 \cdot 21}{2} = 210$.

A single term is the jump the running total makes when that term arrives, so $a_5 = S_5 - S_4$.

$S_5 = 3(25) - 5 = 70, \qquad S_4 = 3(16) - 4 = 44$

Subtract.

$a_5 = 70 - 44 = 26$

In general $a_n = S_n - S_{n-1} = 6n - 4$, which is linear in $n$, so the sequence really is arithmetic (with $d = 6$). The value $70$ is the sum of the first five terms, not the fifth term.

The weekly deposits are $5, 10, 15, \ldots$, arithmetic with $a_1 = 5$ and $d = 5$, so the running total after $n$ weeks is

$S_n = \frac{n}{2}\big(5 + 5n\big) = \frac{5n(n + 1)}{2}.$

Set that equal to the target and simplify.

$\frac{5n(n+1)}{2} = 1050 \;\Rightarrow\; n(n + 1) = 420$

Since $20 \cdot 21 = 420$, the total reaches $1{,}050$ dollars in week $20$. After $19$ weeks it stands at only $950$ dollars.

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

$S_{20} = 10(2a_1 + 19d), \qquad 2S_{10} = 2 \cdot 5(2a_1 + 9d) = 10(2a_1 + 9d)$

Set them equal and cancel the common factor of $10$.

$2a_1 + 19d = 2a_1 + 9d \;\Rightarrow\; 10d = 0 \;\Rightarrow\; d = 0$

So the sequence has to be constant. That is believable: each of the terms $a_{11}$ through $a_{20}$ sits exactly $10d$ above the matching term in the first ten, so the two halves can only be equal when $d = 0$.