Geometric Sequences and Series: Additional Practice (Set 2)

Recover the terms from the sums. The first term is $S_1$, and the second term is what the sum gains on the next step, $S_2 - S_1$.

$S_1 = 3(2 - 1) = 3, \qquad S_2 = 3(4 - 1) = 9 \quad\Longrightarrow\quad a_2 = 9 - 3 = 6$

So $a_1 = 3$ and $r = a_2 / a_1 = 2$. This matches the standard form $S_n = a(r^{\,n} - 1)/(r - 1)$, which for $a = 3$ and $r = 2$ is exactly $3(2^{\,n} - 1)$.

Terms three places apart differ by three multiplications, so $a_7 = a_4 r^3$.

$r^3 = \frac{1458}{54} = 27 \quad\Longrightarrow\quad r = 3$

Stepping back three places from $a_4$ divides by $r^3 = 27$.

$a_1 = \frac{54}{27} = 2$

The sequence runs $2, 6, 18, 54, 162, 486, 1458$.

Multiplying a term by $r$ gives the next one, so multiplying the whole first equation by $r$ turns it into the second: $a_2 + a_3 = r(a_1 + a_2)$.

$36 = r \cdot 12 \quad\Longrightarrow\quad r = 3$

No need to find $a_1$ at all, though it follows easily: $a_1(1 + 3) = 12$ gives $a_1 = 3$, and the sequence $3, 9, 27$ satisfies both conditions.

Write the sum of four terms with an unknown first term $a$.

$S_4 = \frac{a(3^4 - 1)}{3 - 1} = \frac{80a}{2} = 40a$

Setting $40a = 40$ gives $a = 1$. The sequence is $1, 3, 9, 27$, and those four terms do add to $40$.

The ratio is $-3$, so $a = 1$, $r = -3$, and $n = 7$. The exponent $7$ is odd, so $(-3)^7 = -2187$, and subtracting a negative number adds.

$S_7 = \frac{1\bigl(1 - (-3)^7\bigr)}{1 - (-3)} = \frac{1 + 2187}{4} = \frac{2188}{4} = 547$

Checking directly: $1 - 3 + 9 - 27 + 81 - 243 + 729 = 547$.

The summand is geometric with $a = 4$ and $r = \tfrac{1}{2}$, and $k$ runs from $1$ to $5$, so there are five terms.

$\sum_{k=1}^{5} 4\left(\tfrac{1}{2}\right)^{k-1} = \frac{4\left(1 - \left(\tfrac{1}{2}\right)^5\right)}{1 - \tfrac{1}{2}} = 8\left(1 - \tfrac{1}{32}\right) = \frac{31}{4}$

Adding the terms confirms it: $4 + 2 + 1 + \tfrac{1}{2} + \tfrac{1}{4} = \tfrac{31}{4}$. Four terms would give $\tfrac{15}{2}$ and six would give $\tfrac{63}{8}$.

Losing $20$ percent leaves $80$ percent, so the value is multiplied by $r = 0.8$ each year, and the yearly values are geometric.

$24{,}000 \cdot (0.8)^4 = 24{,}000 \cdot 0.4096 = 9830.40$

The car is worth $9830.40$ dollars. Stopping at $(0.8)^3 = 0.512$ gives $12{,}288$ dollars, which is the value after only three years, and $4800$ dollars would be the result of multiplying by $0.2$ once.

With $a = 3$ and $r = 2$ the sum of the first $n$ terms is $S_n = 3(2^{\,n} - 1)$.

$3(2^{\,n} - 1) > 100{,}000 \quad\Longrightarrow\quad 2^{\,n} > 33{,}334.3$

The unknown is an exponent, so take logarithms.

$n > \frac{\log 33{,}334.3}{\log 2} \approx \frac{4.5229}{0.3010} \approx 15.02$

So $n = 16$. Checking the boundary: $S_{15} = 3(32{,}767) = 98{,}301$ falls short, while $S_{16} = 3(65{,}535) = 196{,}605$ clears the target.

Write both sums with the same first term $a$ and ratio $2$, then divide.

$\frac{S_6}{S_3} = \frac{a(2^6 - 1)/(2 - 1)}{a(2^3 - 1)/(2 - 1)} = \frac{63}{7} = 9$

The first term cancels, so the ratio of the sums does not depend on it. The answer $9$ equals $1 + r^3$, which makes sense: the second group of three terms is $r^3 = 8$ times the first group, so together they are $9$ times it.

Consecutive terms share one ratio, so $\dfrac{x+3}{x} = \dfrac{x+9}{x+3}$. Cross-multiplying gives $(x+3)^2 = x(x+9)$.

$x^2 + 6x + 9 = x^2 + 9x \quad\Longrightarrow\quad 9 = 3x \quad\Longrightarrow\quad x = 3$

The quadratic terms cancel, leaving a linear equation with the single solution $x = 3$. The terms are then $3, 6, 12$, with ratio $2$ throughout.

A sequence whose terms are all $6$ has common ratio $r = 1$, and the formula divides by $1 - r = 0$. That division is illegal, so the formula simply does not apply. Return to the definition of the sum.

$S_{20} = 20 \cdot 6 = 120$

The sum is perfectly well defined; it is $na$, twenty copies of $6$. What fails is only the formula, whose derivation divided by $1 - r$ in its final step.

With $a = 3$ and $r = 2$ the sum of the first $n$ terms is $S_n = 3(2^{\,n} - 1)$.

$3(2^{\,n} - 1) = 3069 \quad\Longrightarrow\quad 2^{\,n} - 1 = 1023 \quad\Longrightarrow\quad 2^{\,n} = 1024 = 2^{10}$

So $n = 10$. Ten terms, starting at $3$ and doubling, add to $3069$. The value $1024$ is the power of two, not the number of terms.