Geometric Sequences and Series: Additional Practice (Set 1)

Each term is $\left(\tfrac{1}{2}\right)^{n-1}$, and the last term is $\tfrac{1}{1024} = \left(\tfrac{1}{2}\right)^{10}$, since $2^{10} = 1024$.

$n - 1 = 10 \quad\Longrightarrow\quad n = 11$

The terms are the powers $\left(\tfrac{1}{2}\right)^0$ through $\left(\tfrac{1}{2}\right)^{10}$, which is $11$ of them: one more than the highest exponent.

With $a = 1$ and $r = 2$ the general term is $2^{\,n-1}$, and the last term is $1024 = 2^{10}$, so $n - 1 = 10$ and there are $11$ terms.

$S_{11} = \frac{1(2^{11} - 1)}{2 - 1} = 2048 - 1 = 2047$

A run of doublings starting at $1$ always adds to one less than the next power of two. Counting only $10$ terms would give $1023$.

Identical terms mean the common ratio is $1$, and the sum formula would divide by $1 - r = 0$, so it cannot be used. Add the terms as they stand.

$S_{100} = 100 \cdot (-3) = -300$

The sum is not undefined. When $r = 1$ the sum is $na$, here one hundred copies of $-3$. Only the formula fails, not the sum.

The general term is $a_n = 2 \cdot 3^{\,n-1}$, so the requirement $a_n > 10{,}000$ becomes $3^{\,n-1} > 5000$.

$3^7 = 2187 < 5000, \qquad 3^8 = 6561 > 5000$

So the first exponent that works is $n - 1 = 8$, giving $n = 9$. Checking the boundary: $a_8 = 2 \cdot 2187 = 4374$ is under, while $a_9 = 2 \cdot 6561 = 13{,}122$ is over.

The sixth term is five multiplications past the first, and $5$ is odd, so $(-3)^5 = -243$.

$a_6 = -2 \cdot (-3)^5 = -2 \cdot (-243) = 486$

Two negatives multiply to a positive. The sequence runs $-2, 6, -18, 54, -162, 486$, so the sixth term is indeed positive.

Interest of $6$ percent a year multiplies the balance by $1 + 0.06 = 1.06$ every year, so the yearly balances form a geometric sequence with $r = 1.06$, not $0.06$.

$\text{balance after } 10 \text{ years} = 5000(1.06)^{10}$

Multiplying by $0.06$ would destroy almost all of the money each year. The expression $5000 + 10(0.06)(5000)$ adds the same amount annually, which is simple interest, not compounding.

The sum formula needs the number of terms, so get $n$ from the last term.

$5 \cdot 3^{\,n-1} = 405 \quad\Longrightarrow\quad 3^{\,n-1} = 81 = 3^4 \quad\Longrightarrow\quad n = 5$

Now add the five terms.

$S_5 = \frac{5(3^5 - 1)}{3 - 1} = \frac{5 \cdot 242}{2} = 605$

Checking by hand: $5 + 15 + 45 + 135 + 405 = 605$. Forgetting to divide by $r - 1 = 2$ would give $1210$.

The formula divides by $1 - r$, so it fails exactly when $r = 1$. Identical terms mean $r = 1$, and the denominator becomes zero.

$S_{12} = 12 \cdot 7 = 84$

That sum has to be computed as $na$ instead. The other three have $r = -1$, $r = 2$, and $r = 0.5$, and each gives a nonzero denominator ($2$, $-1$, and $0.5$), so the formula handles them all.

The third term is what the sum gains on the third step: $a_3 = S_3 - S_2 = 21 - 9 = 12$. Writing the terms as $a$, $ar$, and $ar^2$ gives two equations.

$a(1 + r) = 9, \qquad ar^2 = 12$

Substituting $a = 9/(1+r)$ into the second gives $9r^2 = 12(1 + r)$, that is $3r^2 - 4r - 4 = 0$, which factors as $(3r + 2)(r - 2) = 0$.

$r = 2 \quad\text{or}\quad r = -\tfrac{2}{3}$

The problem asks for the positive ratio, so $r = 2$, and the sequence $3, 6, 12$ does give $S_2 = 9$ and $S_3 = 21$.

Each fold doubles the thickness, so the thicknesses form a geometric sequence with $r = 2$ starting at $0.1$ mm, and $20$ folds means $20$ doublings.

$0.1 \cdot 2^{20} = 0.1 \cdot 1{,}048{,}576 = 104{,}857.6 \text{ mm}$

Dividing by $1000$ converts millimetres to metres, giving about $105$ metres, roughly a thirty-storey building. Doubling twenty times multiplies by more than a million, which is why the answer feels impossible.

Terms two places apart differ by two multiplications, so $a_5 = a_3 r^2$.

$r^2 = \frac{48}{12} = 4 \quad\Longrightarrow\quad r = \pm 2$

The ratio is negative, so $r = -2$, and the fourth term is one multiplication past the third.

$a_4 = 12 \cdot (-2) = -24$

The run $12, -24, 48$ has ratio $-2$ throughout, and the two sign changes cancel to leave $a_5$ positive.

With $a = 1$ and $r = 5$ the sum of the first $n$ terms is

$S_n = \frac{1(5^{\,n} - 1)}{5 - 1} = \frac{5^{\,n} - 1}{4} .$

Setting this equal to $781$ gives $5^{\,n} - 1 = 3124$, so $5^{\,n} = 3125 = 5^5$ and $n = 5$. Checking: $1 + 5 + 25 + 125 + 625 = 781$. The value $625$ is the last term, not the number of terms.