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Geometric Sequences and Series: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 A rule with signs

For positive integers $n$ , let $a_n=\frac{(-2)^{n+1}}3$ . Find the common ratio.
1. Hint 1
  Compare the expression at $n+1$ with the expression at $n$ .
2. Hint 2
  All terms are nonzero, so the common coefficient cancels in the quotient.
Answer

$r=-2$ .

Full solution

Divide consecutive terms.
The factors of $\frac13$ cancel, and the exponent in the numerator is one larger.
$\frac{a_{n+1}}{a_n}=-2$
This holds at every index, so the sequence is geometric with ratio $-2$ .

Answer

$r=-2$ .

Key idea

A constant ratio can be found from a formula without listing many terms.
Problem 1: Mark complete
Problem 2 A separated quotient

A geometric sequence has $a_3=6$ and $a_4=-9$ . Find $\frac{a_6}{a_2}$ .
1. Hint 1
  The neighboring known terms determine the ratio, including its sign.
2. Hint 2
  Four multiplications by the ratio connect index two to index six.
Answer

$\frac{81}{16}$ .

Full solution

The ratio is $r=-9/6$ , which is $-\frac32$ .
Every term is nonzero, so the requested quotient is
$\frac{a_6}{a_2}=(-\tfrac32)^4$
giving $\frac{81}{16}$ .
The even number of sign changes makes this quotient positive.

Answer

$\frac{81}{16}$ .

Key idea

The quotient of separated geometric terms is the common ratio raised to their index gap.
Problem 2: Mark complete
Problem 3 A sum across zero

Evaluate $\sum_{k=-2}^1 3\cdot2^k$ .
1. Hint 1
  The four terms are geometric even though the lower index is negative.
2. Hint 2
  The first included term is $3\cdot2^{-2}$ and the ratio is $2$ .
Answer

$\frac{45}{4}$ .

Full solution

The four terms are $\frac34,\frac32,3,6$ .
With $a=\frac34$ , $r=2\ne1$ , and $n=4$ ,
$S=\frac34(2^4-1)$
which is $\frac{45}{4}$ .
Direct addition gives numerators $3+6+12+24=45$ over $4$ .

Answer

$\frac{45}{4}$ .

Key idea

Finite geometric sums use the first included term, even when its exponent is negative.
Problem 3: Mark complete
Problem 4 A compressed file record

A file occupies $1000$ kilobytes at stage $1$ . Each later stage retains $80\%$ of the previous stage's size. Find the first stage whose file is smaller than $300$ kilobytes and give its size to the nearest tenth of a kilobyte.
1. Hint 1
  The stage-one value is the seed, so stage n has undergone n minus one reductions.
2. Hint 2
  Solve $0.8^{n-1}<0.3$ and choose the smallest allowed integer index.
Answer

Stage $7$ ; size $\approx262.1$ kilobytes.

Full solution

The model is
$a_n=1000(0.8)^{n-1}.$
The threshold is
$0.8^{n-1}<0.3.$
Since $\ln(0.8)<0$ , dividing reverses the logarithmic inequality.
$n>1+\frac{\ln(0.3)}{\ln(0.8)}$

The smallest integer that satisfies this is $7$ .
At stage six the size is $327.68$ kilobytes, still above the threshold; at stage seven it is $262.144$ kilobytes, which rounds to $262.1$ .

Answer

Stage $7$ ; size $\approx262.1$ kilobytes.

Key idea

A geometric threshold must be converted to the correct integer term index after solving the logarithmic inequality.
Problem 4: Mark complete
Problem 5 Two signed terms

A geometric sequence has $a_2=-12$ and $a_5=324$ . Find its common ratio, first term, and the sum of its first four terms.
1. Hint 1
  Dividing the fifth term by the second gives the cube of the ratio.
2. Hint 2
  After finding the real cube root, work backward one step to the first term.
Answer

$r=-3$ , $a_1=4$ , $S_4=-80$ .

Full solution

The three-step quotient gives $r^3=324/(-12)=-27$ , so $r=-3$ .
Since this is nonzero, $a_1=(-12)/(-3)$ , which is $4$ .

The first four terms are $4,-12,36,-108$ .
Their sum is
$S_4=-80$
The next term is $324$ , matching the second given term.

Answer

$r=-3$ , $a_1=4$ , $S_4=-80$ .

Key idea

An odd index gap determines the sign of a real geometric ratio as well as its size.
Problem 5: Mark complete
Problem 6 Three deposits

An account starts empty. At the beginning of each of three months, $40$ dollars is deposited, and at the end of each month the whole balance grows by $10\%$ . There are no other transactions. Find the balance just after the third month's growth, and write it as a finite geometric sum.
1. Hint 1
  Different deposits receive different numbers of growth steps.
2. Hint 2
  Count the growth steps received by the earliest and by the latest deposit.
Answer

$145.64$ dollars; $40\sum_{j=1}^{3}(1.1)^j$ dollars.

Full solution

The first deposit grows three times, the second twice and the third once, so they end at $40(1.1)^3=53.24$ , $40(1.1)^2=48.40$ and $40(1.1)=44$ dollars.

The balance is $40\sum_{j=1}^{3}(1.1)^j$ , a geometric sum with first term $44$ and ratio $1.1$ :
$\frac{44((1.1)^3-1)}{1.1-1}=145.64$
dollars, matching $53.24+48.40+44$ .

Answer

$145.64$ dollars; $40\sum_{j=1}^{3}(1.1)^j$ dollars.

Key idea

With equal monthly deposits and a fixed monthly growth factor, the deposits' final contributions form a geometric sequence.
Problem 6: Mark complete
Problem 7 The surviving entries

A finite geometric sequence starts at $11$ and has common ratio $4$ . A record subtracts its sum $S$ from $4S$ and cancels every shared term. The only terms left are $11\cdot4^5$ and $-11$ . Recover the number of terms, the last term, and $S$ . Explain why the shared terms cancel and why the positive surviving term is not the original last term.
1. Hint 1
  Multiplying the sum by the common ratio moves every term to the next position in the sequence.
2. Hint 2
  Compare the exponent on the last term of $S$ with the exponent on the last term of $4S$ .
3. Hint 3
  The recorded subtraction is $4S-S$ ; combine its surviving terms to recover the original sum.
Answer

$5$ terms; last term $2816$ ; $S=3751$ ; the positive survivor is $4a_5$ , one ratio step beyond the last term.

Full solution

If the sum has $n$ terms, their powers of $4$ run from exponent $0$ through exponent $n-1$ .
The powers in $4S$ run from exponent $1$ through exponent $n$ .
Subtracting $S$ from $4S$ cancels every term with exponent $1$ through $n-1$ , leaving
$4S-S=11\cdot4^n-11$
The recorded positive term therefore gives $n=5$ .

That positive surviving term is four times the original last term.
The last term itself is
$\begin{aligned} a_5&=11\cdot4^4 \\ &=2816. \end{aligned}$

The recorded subtraction gives
$\begin{aligned} 3S&=11\cdot4^5-11 \\ &=11253 \\ S&=3751. \end{aligned}$
For a direct check, the terms are $11,44,176,704,2816$ , whose total is $3751$ .

Answer

$5$ terms; last term $2816$ ; $S=3751$ ; the positive survivor is $4a_5$ , one ratio step beyond the last term.

Key idea

The uncanceled final term in a shifted geometric sum is one ratio beyond the original last term.
Problem 7: Mark complete
Problem 8 A growth claim

A geometric sequence starts at $a_1=-5$ with ratio $r=2$ . A student says the terms grow in size but decrease in value. Is that description correct? Explain.
1. Hint 1
  Size refers to absolute value, while numerical order uses the sign.
2. Hint 2
  Compare the first three terms and their absolute values.
Answer

Yes; the values decrease and their absolute values increase.

Full solution

The terms begin $-5,-10,-20$ .
Each is less than the previous term, but their sizes are $5,10,20$ .
In general
$|a_{n+1}|=2|a_n|$
while multiplying a negative value by two makes it smaller in numerical order.
The distinction holds at every step.

Answer

Yes; the values decrease and their absolute values increase.

Key idea

Growth in absolute value need not mean growth in numerical value.
Problem 8: Mark complete
Problem 9 Two matching totals

A geometric sequence has positive terms. Its first two terms total $26$ , and its next two terms also total $26$ . A student says these totals are insufficient to determine the sum of its first $11$ terms. Is the student correct? Find that sum if it is determined.
1. Hint 1
  Compare one two-term block with the block immediately after it.
2. Hint 2
  Advancing a whole two-term block multiplies its total by $r^2$ ; use positivity before recovering the first term.
Answer

No; the first $11$ terms total $143$ .

Full solution

Let the first term be $a>0$ and the positive ratio be $r$ .
The first block is $a(1+r)=26$ , and the next block is
$ar^2(1+r)=26.$
Dividing these nonzero totals gives
$r^2=1$
Positivity selects $r=1$ .

Now $2a=26$ , so every term is $13$ .
The sum of eleven constant terms is
$11\cdot13=143.$
The information therefore determines the total; the quotient formula with denominator $1-r$ is unnecessary in this constant case.

Answer

No; the first $11$ terms total $143$ .

Key idea

Equal nonzero geometric block totals can determine the ratio even when no individual term is given.
Problem 9: Mark complete
Problem 10 A proposed term rule

A geometric sequence has $a_1=7$ and $r=\frac13$ . A student proposes $a_n=21(\frac13)^n$ for $n\ge1$ . Is the proposed rule correct? Justify it by checking the seed and the recurrence.
1. Hint 1
  The coefficient need not equal the first term if the exponent starts at n instead of n minus one.
2. Hint 2
  Check the value at index one, then compare successive formula values.
Answer

Yes; it gives the seed $7$ and the common ratio $\frac13$ .

Full solution

At index one the proposed formula gives $21/3=7$ , matching the seed.
Consecutive values have ratio $\frac13$ , since the exponent increases by one.
Equivalently,
$21(\tfrac13)^n=7(\tfrac13)^{n-1}$
for each positive integer $n$ , which is the standard term rule.

Answer

Yes; it gives the seed $7$ and the common ratio $\frac13$ .

Key idea

An alternative geometric formula is valid when its seed and its step multiplier both agree.
Problem 10: Mark complete

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