Sequences and Series Practice Questions

In an arithmetic sequence, each term is found by adding a constant value called the common difference to the previous term. Find the common difference by subtracting consecutive terms:

\(7 – 3 = 4\)

The common difference is 4. Add it to the last given term to find the next term:

\(19 + 4 = 23\)

In a geometric sequence, each term is found by multiplying the previous term by a constant value called the common ratio. Find the common ratio by dividing consecutive terms:

\(\large{\frac{6}{2}}\) \(\:= 3\)

The common ratio is 3. Multiply the last given term by the common ratio to find the next term:

\(54 \times 3 = 162\)

The formula for the nth term of an arithmetic sequence is:

\(a_n = a_1 + (n – 1)d\)

where \(a_1\) is the first term and \(d\) is the common difference. Here, \(a_1 = 5\) and \(d = 4\). Substitute and simplify:

\(a_n = 5 + (n – 1)(4)\)\(\:= 5 + 4n – 4\)\(\:= 4n + 1\)

You can verify:

• \(a_1 = 4(1) + 1 = 5\) ✓
• \(a_2 = 4(2) + 1 = 9\) ✓

The formula for the nth term of a geometric sequence is:

\(a_n = a_1 \cdot r^{n-1}\)

where \(a_1\) is the first term and \(r\) is the common ratio. Substitute \(a_1 = 3\), \(r = 2\), and \(n = 8\):

\(a_8 = 3 \cdot 2^{8-1} = 3 \cdot 2^7 = 3 \cdot 128 = 384\)

Choice B uses \(2^8 = 256\) instead of \(2^7\), a common error that comes from forgetting to subtract 1 from the exponent.

Check for a common difference (arithmetic) by subtracting consecutive terms:

\(12 – 4 = 8\), \(\:36 – 12 = 24\), \(\:108 – 36 = 72\)

The differences are not constant, so the sequence is not arithmetic.

Now check for a common ratio (geometric) by dividing consecutive terms:

\(\large{\frac{12}{4}}\) \(\:= 3\), \(\:\large{\frac{36}{12}}\) \(\:= 3\), \(\:\large{\frac{108}{36}}\) \(\:= 3\)

The ratio is constant at 3, so this is a geometric sequence with common ratio \(r = 3\).

The sum of the first \(n\) terms of an arithmetic series is:

\(S_n = \large{\frac{n}{2}}\)\((a_1 + a_n)\)

First, find the 20th term using the nth term formula:

\(a_{20} = 3 + (20 – 1)(5)\)\(\:= 3 + 95 = 98\)

Now substitute into the sum formula with \(n = 20\), \(a_1 = 3\), and \(a_{20} = 98\):

\(S_{20} = \large{\frac{20}{2}}\)\((3 + 98) = 10(101) = 1{,}010\)

The sum of the first \(n\) terms of a geometric series (when \(r \neq 1\)) is:

\(S_n = a_1 \cdot \dfrac{1 – r^n}{1 – r}\)

Substitute \(a_1 = 4\), \(r = 3\), and \(n = 6\):

\(S_6 = 4 \cdot \large{\frac{1 – 3^6}{1 – 3}}\) \(\:= 4 \cdot \large{\frac{1 – 729}{-2}}\) \(\:= 4 \cdot \large{\frac{-728}{-2}}\) \(\:= 4 \cdot 364 = 1{,}456\)

First, find the common ratio: \(\frac{8}{16} = \frac{1}{2}\). Since \(|r| \lt 1\), the infinite series converges and has a finite sum. The formula for the sum of an infinite geometric series is:

\(S = \dfrac{a_1}{1 – r}\)

Substitute \(a_1 = 16\) and \(r = \frac{1}{2}\):

\(S = \dfrac{16}{1 – \frac{1}{2}} = \dfrac{16}{\frac{1}{2}} = 32\)

If \(|r| \geq 1\), the series diverges and has no finite sum — in that case, choice D would be correct.

Sigma notation tells you to substitute each integer value of \(k\) from 1 to 5 into the expression \(2k + 1\) and add the results:

• \(k = 1{:}\; 2(1) + 1 = 3\)
• \(k = 2{:}\; 2(2) + 1 = 5\)
• \(k = 3{:}\; 2(3) + 1 = 7\)
• \(k = 4{:}\; 2(4) + 1 = 9\)
• \(k = 5{:}\; 2(5) + 1 = 11\)

Now add all five terms:

\(3 + 5 + 7 + 9 + 11 = 35\)

A recursive formula defines each term using the previous term. Start with \(a_1 = 2\) and apply the rule \(a_n = 3a_{n-1} – 1\) repeatedly:

\(a_2 = 3(2) – 1 = 5\)

\(a_3 = 3(5) – 1 = 14\)

\(a_4 = 3(14) – 1 = 41\)

Choice A gives \(a_3\) instead of \(a_4\) — be sure to continue computing until you reach the requested term.

Use the nth term formula \(a_n = a_1 + (n – 1)d\) to set up two equations:

\(a_3 = a_1 + 2d = 14\)

\(a_7 = a_1 + 6d = 30\)

Subtract the first equation from the second to eliminate \(a_1\):

\(4d = 16 \implies d = 4\)

Substitute \(d = 4\) back into the first equation:

\(a_1 + 2(4) = 14 \implies a_1 = 6\)

Each term in the series is a multiple of 5: the 1st term is \(5(1)\), the 2nd is \(5(2)\), and so on through \(5(5)\). This pattern matches \(5k\) where \(k\) goes from 1 to 5.

Choice C starts at \(k = 0\), which would include an extra term of \(5(0) = 0\) — it gives the correct sum by coincidence, but the series would have 6 terms instead of 5. Choice D represents \(5^1 + 5^2 + 5^3 + 5^4 + 5^5\), which is an entirely different series.

The weekly earnings form an arithmetic sequence with \(a_1 = 200\) and \(d = 15\). First, find the earnings in week 10:

\(a_{10} = 200 + (10 – 1)(15)\)\(\:= 200 + 135 = 335\)

Use the arithmetic series sum formula to find the total earnings:

\(S_{10} = \large{\frac{10}{2}(200 + 335)}\) \(\:= 5(535) = 2{,}675\)

The worker earns $2,675 in total over the first 10 weeks.

First, find the common ratio: \(\frac{10}{5} = 2\). An infinite geometric series only converges when \(|r| \lt 1\). Since \(|2| = 2\), which is not less than 1, the series diverges.

This means the partial sums grow without bound and the series has no finite sum. The sum formula \(S = \frac{a_1}{1 – r}\) cannot be applied when \(|r| \geq 1\).

Substitute each integer value of \(k\) from 1 to 4 into the expression \(3 \cdot 2^k\):

• \(k = 1{:}\; 3 \cdot 2^1 = 6\)
• \(k = 2{:}\; 3 \cdot 2^2 = 12\)
• \(k = 3{:}\; 3 \cdot 2^3 = 24\)
• \(k = 4{:}\; 3 \cdot 2^4 = 48\)

Now add all four terms:

\(6 + 12 + 24 + 48 = 90\)

Choice B gives only the last term (\(k = 4\)) rather than the sum of all four terms.